How do you determine if the following series converges?












3












$begingroup$


$$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{2}}$$
I tried using the limit comparison test with $$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{}}$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of
$mathrm{e}^x$, but I don't know where else to start. Any suggestions?






convergence







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$endgroup$












  • $begingroup$
    This is susceptible to the same approach as my answer to a different question.
    $endgroup$
    – T. Bongers
    39 mins ago










  • $begingroup$
    Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
    $endgroup$
    – JavaMan
    33 mins ago












  • $begingroup$
    I cannot believe how horrible my intuition is with this stuff especially given how old I am.
    $endgroup$
    – Randall
    3 mins ago
















3












$begingroup$


$$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{2}}$$
I tried using the limit comparison test with $$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{}}$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of
$mathrm{e}^x$, but I don't know where else to start. Any suggestions?






convergence







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is susceptible to the same approach as my answer to a different question.
    $endgroup$
    – T. Bongers
    39 mins ago










  • $begingroup$
    Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
    $endgroup$
    – JavaMan
    33 mins ago












  • $begingroup$
    I cannot believe how horrible my intuition is with this stuff especially given how old I am.
    $endgroup$
    – Randall
    3 mins ago














3












3








3





$begingroup$


$$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{2}}$$
I tried using the limit comparison test with $$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{}}$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of
$mathrm{e}^x$, but I don't know where else to start. Any suggestions?






convergence







share|cite|improve this question









$endgroup$




$$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{2}}$$
I tried using the limit comparison test with $$sum_{k=1}^infty mathrm{(1-frac{1}{k})}^{mathrm{k}^{}}$$ but this leads to a limit of 0, which doesn't help. I think this may involve some use of
$mathrm{e}^x$, but I don't know where else to start. Any suggestions?






convergence




convergence






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asked 49 mins ago









JayJay

334




334












  • $begingroup$
    This is susceptible to the same approach as my answer to a different question.
    $endgroup$
    – T. Bongers
    39 mins ago










  • $begingroup$
    Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
    $endgroup$
    – JavaMan
    33 mins ago












  • $begingroup$
    I cannot believe how horrible my intuition is with this stuff especially given how old I am.
    $endgroup$
    – Randall
    3 mins ago


















  • $begingroup$
    This is susceptible to the same approach as my answer to a different question.
    $endgroup$
    – T. Bongers
    39 mins ago










  • $begingroup$
    Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
    $endgroup$
    – JavaMan
    33 mins ago












  • $begingroup$
    I cannot believe how horrible my intuition is with this stuff especially given how old I am.
    $endgroup$
    – Randall
    3 mins ago
















$begingroup$
This is susceptible to the same approach as my answer to a different question.
$endgroup$
– T. Bongers
39 mins ago




$begingroup$
This is susceptible to the same approach as my answer to a different question.
$endgroup$
– T. Bongers
39 mins ago












$begingroup$
Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
$endgroup$
– JavaMan
33 mins ago






$begingroup$
Do you know what $left(1-frac{1}{k}right)^{k}$ converges to?
$endgroup$
– JavaMan
33 mins ago














$begingroup$
I cannot believe how horrible my intuition is with this stuff especially given how old I am.
$endgroup$
– Randall
3 mins ago




$begingroup$
I cannot believe how horrible my intuition is with this stuff especially given how old I am.
$endgroup$
– Randall
3 mins ago










4 Answers
4






active

oldest

votes


















0












$begingroup$

HINT:



Note that $$left( 1-frac1k right)^kle e^{-1}$$



Can you finish?






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    The root test works. Consider
    $$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
    hence the series converges.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      The ratio test is also interesting
      $$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
      $$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$



      Develop as a Taylor series for large values of $k$ to get
      $$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+O(left(frac{1}{k^3}right)$$
      Continue with Taylor
      $$frac{a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 eleft(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e$$






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        $begin{array}\
        (1-frac1{k})^{k^2}
        &=(frac{k-1}{k})^{k^2}\
        &=dfrac1{(frac{k}{k-1})^{k^2}}\
        &=dfrac1{(1+frac{1}{k-1})^{k^2}}\
        &=dfrac1{((1+frac{1}{k-1})^{k})^k}\
        &<dfrac1{(1+frac{k}{k-1})^k}
        qquadtext{by Bernoulli}\
        &=dfrac1{(frac{2k-1}{k-1})^k}\
        &<dfrac1{(frac{2k-2}{k-1})^k}\
        &=dfrac1{2^k}\
        end{array}
        $



        and the sum of this converges.






        share|cite|improve this answer









        $endgroup$













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          4 Answers
          4






          active

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          4 Answers
          4






          active

          oldest

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          active

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          active

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          0












          $begingroup$

          HINT:



          Note that $$left( 1-frac1k right)^kle e^{-1}$$



          Can you finish?






          share|cite|improve this answer









          $endgroup$


















            0












            $begingroup$

            HINT:



            Note that $$left( 1-frac1k right)^kle e^{-1}$$



            Can you finish?






            share|cite|improve this answer









            $endgroup$
















              0












              0








              0





              $begingroup$

              HINT:



              Note that $$left( 1-frac1k right)^kle e^{-1}$$



              Can you finish?






              share|cite|improve this answer









              $endgroup$



              HINT:



              Note that $$left( 1-frac1k right)^kle e^{-1}$$



              Can you finish?







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 35 mins ago









              Mark ViolaMark Viola

              132k1277174




              132k1277174























                  3












                  $begingroup$

                  The root test works. Consider
                  $$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
                  hence the series converges.






                  share|cite|improve this answer









                  $endgroup$


















                    3












                    $begingroup$

                    The root test works. Consider
                    $$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
                    hence the series converges.






                    share|cite|improve this answer









                    $endgroup$
















                      3












                      3








                      3





                      $begingroup$

                      The root test works. Consider
                      $$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
                      hence the series converges.






                      share|cite|improve this answer









                      $endgroup$



                      The root test works. Consider
                      $$lim sup sqrt[k]{left(1 - frac{1}{k}right)^{k^2}} = lim sup left(1 - frac{1}{k}right)^k = e^{-1} < 1,$$
                      hence the series converges.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 38 mins ago









                      Theo BenditTheo Bendit

                      18.8k12253




                      18.8k12253























                          0












                          $begingroup$

                          The ratio test is also interesting
                          $$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
                          $$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$



                          Develop as a Taylor series for large values of $k$ to get
                          $$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+O(left(frac{1}{k^3}right)$$
                          Continue with Taylor
                          $$frac{a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 eleft(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e$$






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            The ratio test is also interesting
                            $$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
                            $$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$



                            Develop as a Taylor series for large values of $k$ to get
                            $$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+O(left(frac{1}{k^3}right)$$
                            Continue with Taylor
                            $$frac{a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 eleft(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e$$






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              The ratio test is also interesting
                              $$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
                              $$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$



                              Develop as a Taylor series for large values of $k$ to get
                              $$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+O(left(frac{1}{k^3}right)$$
                              Continue with Taylor
                              $$frac{a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 eleft(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e$$






                              share|cite|improve this answer









                              $endgroup$



                              The ratio test is also interesting
                              $$a_k=left(1-frac{1}{k}right)^{k^2}implies log(a_k)=k ^2 logleft(1-frac{1}{k}right)$$
                              $$log(a_{k+1})-log(a_k)=(k+1) ^2 logleft(1-frac{1}{k+1}right)-k ^2 logleft(1-frac{1}{k}right)$$



                              Develop as a Taylor series for large values of $k$ to get
                              $$log(a_{k+1})-log(a_k)=-1+frac{1}{3 k^2}+O(left(frac{1}{k^3}right)$$
                              Continue with Taylor
                              $$frac{a_{k+1}}{a_k}=e^{log(a_{k+1})-log(a_k)}=frac 1 eleft(1+frac{1}{3 k^2}+Oleft(frac{1}{k^3}right)right)to frac 1 e$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 25 mins ago









                              Claude LeiboviciClaude Leibovici

                              122k1157134




                              122k1157134























                                  0












                                  $begingroup$

                                  $begin{array}\
                                  (1-frac1{k})^{k^2}
                                  &=(frac{k-1}{k})^{k^2}\
                                  &=dfrac1{(frac{k}{k-1})^{k^2}}\
                                  &=dfrac1{(1+frac{1}{k-1})^{k^2}}\
                                  &=dfrac1{((1+frac{1}{k-1})^{k})^k}\
                                  &<dfrac1{(1+frac{k}{k-1})^k}
                                  qquadtext{by Bernoulli}\
                                  &=dfrac1{(frac{2k-1}{k-1})^k}\
                                  &<dfrac1{(frac{2k-2}{k-1})^k}\
                                  &=dfrac1{2^k}\
                                  end{array}
                                  $



                                  and the sum of this converges.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    $begin{array}\
                                    (1-frac1{k})^{k^2}
                                    &=(frac{k-1}{k})^{k^2}\
                                    &=dfrac1{(frac{k}{k-1})^{k^2}}\
                                    &=dfrac1{(1+frac{1}{k-1})^{k^2}}\
                                    &=dfrac1{((1+frac{1}{k-1})^{k})^k}\
                                    &<dfrac1{(1+frac{k}{k-1})^k}
                                    qquadtext{by Bernoulli}\
                                    &=dfrac1{(frac{2k-1}{k-1})^k}\
                                    &<dfrac1{(frac{2k-2}{k-1})^k}\
                                    &=dfrac1{2^k}\
                                    end{array}
                                    $



                                    and the sum of this converges.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      $begin{array}\
                                      (1-frac1{k})^{k^2}
                                      &=(frac{k-1}{k})^{k^2}\
                                      &=dfrac1{(frac{k}{k-1})^{k^2}}\
                                      &=dfrac1{(1+frac{1}{k-1})^{k^2}}\
                                      &=dfrac1{((1+frac{1}{k-1})^{k})^k}\
                                      &<dfrac1{(1+frac{k}{k-1})^k}
                                      qquadtext{by Bernoulli}\
                                      &=dfrac1{(frac{2k-1}{k-1})^k}\
                                      &<dfrac1{(frac{2k-2}{k-1})^k}\
                                      &=dfrac1{2^k}\
                                      end{array}
                                      $



                                      and the sum of this converges.






                                      share|cite|improve this answer









                                      $endgroup$



                                      $begin{array}\
                                      (1-frac1{k})^{k^2}
                                      &=(frac{k-1}{k})^{k^2}\
                                      &=dfrac1{(frac{k}{k-1})^{k^2}}\
                                      &=dfrac1{(1+frac{1}{k-1})^{k^2}}\
                                      &=dfrac1{((1+frac{1}{k-1})^{k})^k}\
                                      &<dfrac1{(1+frac{k}{k-1})^k}
                                      qquadtext{by Bernoulli}\
                                      &=dfrac1{(frac{2k-1}{k-1})^k}\
                                      &<dfrac1{(frac{2k-2}{k-1})^k}\
                                      &=dfrac1{2^k}\
                                      end{array}
                                      $



                                      and the sum of this converges.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 14 mins ago









                                      marty cohenmarty cohen

                                      73.8k549128




                                      73.8k549128






























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                                          この記事は検証可能な参考文献や出典が全く示されていないか、不十分です。 出典を追加して記事の信頼性向上にご協力ください。 （ 2019年1月 ） 「 サソリ 」のその他の用法については「サソリ (曖昧さ回避)」をご覧ください。 「 スコーピオン 」と「 スコルピオン 」はこの項目へ転送されています。その他の用法については「スコーピオン」をご覧ください。 サソリ目 生息年代: 430–0 Ma PreЄ Є O S D C P T J K Pg N シルル紀前期-現世 ダイオウサソリ（ Pandinus Imperator ） 分類 界 : 動物界 Animalia 門 : 節足動物門 Arthropoda 亜門 : 鋏角亜門 Chelicerata 綱 : クモガタ綱 Arachnida 目 : サソリ目 Scorpiones 学名 Scorpiones C. L. Koch, 1837 英名 Scorpion 上科 Pseudochactoidea キョクトウサソリ上科 Buthoidea Chaeriloidea Chactoidea Iuroidea Scorpionoidea サソリ （ サソリ類 、 蠍 、 蝎 、 scorpion ）は、鋏角亜門・クモガタ綱の サソリ目 （ Scorpiones ）に属する節足動物の総称である。体の前端に鋏型の触肢、後端に毒針を有する捕食者である。 1700種以上を含め、最古の化石記録は4億3千万年前のシルル紀まで遡る。有毒生物として広く知られるが、人間の命にかかわる毒を持つものはごく一部の種類に過ぎない。 目次 1 形態 1.1 前体 1.2 後体 1.2.1 中体 1.2.2 終体 2 雌雄 3 生態 3.1 天敵 3.2 生殖 4 分布 5 毒 6 蛍光現象 7 分類
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                                          広島県道265号伴広島線

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                                          一般県道 広島県道265号伴広島線 路線延長 km 起点 広島県広島市安佐南区伴東7丁目 終点 広島県広島市中区榎町 接続する 主な道路 （記法） 広島県道38号広島豊平線 広島県道71号広島湯来線 山陽自動車道五日市IC 国道183号（ 国道261号重複） ■テンプレート（■ノート ■使い方） ■PJ道路 広島県道265号伴広島線 （ひろしまけんどう265ごう ともひろしません）は広島県広島市安佐南区から広島県広島市中区を結ぶ一般県道である。 目次 1 路線データ 2 通過する自治体 3 見どころ 4 接続道路 4.1 本線 4.2 旧道 5 別名 6 関連項目 路線データ 本線 起点：広島市安佐南区伴東7丁目・大原下橋南詰交差点（広島県道38号広島豊平線（本線・旧道）交点） 終点：広島市中区榎町・十日市交差点（国道183号（国道261号重用）交点） 総延長： 旧道 起点：広島市安佐南区伴東4丁目・新細坂交差点（広島県道38号広島豊平線交点） 終点：広島市安佐南区伴中央4丁目・寺組橋(南)交差点（広島県道265号伴広島線（本線）交点） 総延長： 通過する自治体 広島県 広島市（安佐南区、佐伯区、西区、中区） 見どころ 広島高速交通広島新交通1号線（アストラムライン） 広島広域公園 広島広域公園陸上競技場（広島ビッグアーチ） 己斐峠 己斐橋（太田川放水路） 天満橋（天満川） 接続道路 本線 広島県道38号広島豊平線（本線・旧道）（広島市安佐南区伴東7丁目・大原下橋南詰交差点〔起点〕） 広島県道71号広島湯来線（本線）（広島市安佐南区伴中央2丁目・伴交番入口交差点 - 広島市安佐南区伴中央4丁目・寺組橋(南)交差点間で重用） 広島県道71号広島湯来線（旧道）（広島県道265号伴広島線（旧道）重用）（広島市安佐南区伴中央4丁目・寺組橋(南)交差点） 広島県道71号広島湯来線（広島市安佐南区伴中央4丁目・寺組橋(南)交差点 - 広島市佐伯区五日市町石内・五月が丘(南)交差点間で重用）
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                                          Accessing regular linux commands in Huawei's Dopra Linux

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                                          4 Backstory: I have a Huawei HG8245 router in my house and I want to change the default username/password. I've tried following this guide but the config file always gets reset. My ISP isn't helping either so I'm looking for other ways to accomplish this. The router has telnet access so this seems like a way to change the credentials. The first shell I gain access to looks like this: WAP> Pressing "?" gives me a list of available commands which I've pasted here, the two interesting ones are "su" and "shell". I can't seem to get su access, even after logging in as root (challenge password prompt, and then nothing happens), but shell gives me a proper linux shell. WAP>shell BusyBox v1.18.4 (2017-12-26 17:06:34 CST) built-in shell (ash) Enter 'hel
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