Sum to infinite series












2














$S=1+4/7+9/49+16/343+....$



I've to find S when the number of terms go to infinity.



We are told just to find sums of AP and GP series. How can we find this?






sequences-and-series







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  • 5




    Are you sure about the $16/243$ term? It would be more likely to be $16/color{red}{3}43$.
    – mrtaurho
    40 mins ago










  • I'm sorry. I wrote it wrong.
    – Ashish Yadav
    37 mins ago
















2














$S=1+4/7+9/49+16/343+....$



I've to find S when the number of terms go to infinity.



We are told just to find sums of AP and GP series. How can we find this?






sequences-and-series







share|cite|improve this question









New contributor




Ashish Yadav is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 5




    Are you sure about the $16/243$ term? It would be more likely to be $16/color{red}{3}43$.
    – mrtaurho
    40 mins ago










  • I'm sorry. I wrote it wrong.
    – Ashish Yadav
    37 mins ago














2












2








2







$S=1+4/7+9/49+16/343+....$



I've to find S when the number of terms go to infinity.



We are told just to find sums of AP and GP series. How can we find this?






sequences-and-series







share|cite|improve this question









New contributor




Ashish Yadav is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











$S=1+4/7+9/49+16/343+....$



I've to find S when the number of terms go to infinity.



We are told just to find sums of AP and GP series. How can we find this?






sequences-and-series




sequences-and-series






share|cite|improve this question









New contributor




Ashish Yadav is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Ashish Yadav is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 37 mins ago





















New contributor




Ashish Yadav is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 43 mins ago









Ashish Yadav

133




133




New contributor




Ashish Yadav is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Ashish Yadav is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Ashish Yadav is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 5




    Are you sure about the $16/243$ term? It would be more likely to be $16/color{red}{3}43$.
    – mrtaurho
    40 mins ago










  • I'm sorry. I wrote it wrong.
    – Ashish Yadav
    37 mins ago














  • 5




    Are you sure about the $16/243$ term? It would be more likely to be $16/color{red}{3}43$.
    – mrtaurho
    40 mins ago










  • I'm sorry. I wrote it wrong.
    – Ashish Yadav
    37 mins ago








5




5




Are you sure about the $16/243$ term? It would be more likely to be $16/color{red}{3}43$.
– mrtaurho
40 mins ago




Are you sure about the $16/243$ term? It would be more likely to be $16/color{red}{3}43$.
– mrtaurho
40 mins ago












I'm sorry. I wrote it wrong.
– Ashish Yadav
37 mins ago




I'm sorry. I wrote it wrong.
– Ashish Yadav
37 mins ago










1 Answer
1






active

oldest

votes


















8














The $n$th term of the sequence is $T_n=frac{n^2}{7^{n-1}}$.
To see that the series converges,
$$r=frac{a_{n+1}}{a_n}=frac{(n+1)^2cdot7^{n-1}}{7^ncdot n^2}$$
$$=frac{(1+1/n)^2}{7}$$
As n tends to $infty$, $r$ tends to $1/7$. Since $r<1$, the series converges. (Thanks to DonAntonio for this :) )



Note that
$$S=1+frac{4}{7}+frac{9}{7^2}+frac{16}{7^3}cdots$$
$$S-S/7=6S/7=1+frac{3}{7}+frac{5}{7^2}+frac{7}{7^3}cdots$$
$$6S/7-6S/49=1+frac{2}{7}+frac{2}{7^2}+frac{2}{7^3}cdots$$
$$frac{36S}{49}=1+frac{2}{7}cdotfrac{7}{6}=frac{4}{3}$$
$$implies S=frac{49}{27}$$






share|cite|improve this answer



















  • 2




    Nice...but it must first be established the first series converges
    – DonAntonio
    35 mins ago








  • 1




    Thanks for solution sir.
    – Ashish Yadav
    35 mins ago






  • 1




    @DonAntonio editted. Thanks!
    – Ankit Kumar
    29 mins ago











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1 Answer
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active

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votes








1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









8














The $n$th term of the sequence is $T_n=frac{n^2}{7^{n-1}}$.
To see that the series converges,
$$r=frac{a_{n+1}}{a_n}=frac{(n+1)^2cdot7^{n-1}}{7^ncdot n^2}$$
$$=frac{(1+1/n)^2}{7}$$
As n tends to $infty$, $r$ tends to $1/7$. Since $r<1$, the series converges. (Thanks to DonAntonio for this :) )



Note that
$$S=1+frac{4}{7}+frac{9}{7^2}+frac{16}{7^3}cdots$$
$$S-S/7=6S/7=1+frac{3}{7}+frac{5}{7^2}+frac{7}{7^3}cdots$$
$$6S/7-6S/49=1+frac{2}{7}+frac{2}{7^2}+frac{2}{7^3}cdots$$
$$frac{36S}{49}=1+frac{2}{7}cdotfrac{7}{6}=frac{4}{3}$$
$$implies S=frac{49}{27}$$






share|cite|improve this answer



















  • 2




    Nice...but it must first be established the first series converges
    – DonAntonio
    35 mins ago








  • 1




    Thanks for solution sir.
    – Ashish Yadav
    35 mins ago






  • 1




    @DonAntonio editted. Thanks!
    – Ankit Kumar
    29 mins ago
















8














The $n$th term of the sequence is $T_n=frac{n^2}{7^{n-1}}$.
To see that the series converges,
$$r=frac{a_{n+1}}{a_n}=frac{(n+1)^2cdot7^{n-1}}{7^ncdot n^2}$$
$$=frac{(1+1/n)^2}{7}$$
As n tends to $infty$, $r$ tends to $1/7$. Since $r<1$, the series converges. (Thanks to DonAntonio for this :) )



Note that
$$S=1+frac{4}{7}+frac{9}{7^2}+frac{16}{7^3}cdots$$
$$S-S/7=6S/7=1+frac{3}{7}+frac{5}{7^2}+frac{7}{7^3}cdots$$
$$6S/7-6S/49=1+frac{2}{7}+frac{2}{7^2}+frac{2}{7^3}cdots$$
$$frac{36S}{49}=1+frac{2}{7}cdotfrac{7}{6}=frac{4}{3}$$
$$implies S=frac{49}{27}$$






share|cite|improve this answer



















  • 2




    Nice...but it must first be established the first series converges
    – DonAntonio
    35 mins ago








  • 1




    Thanks for solution sir.
    – Ashish Yadav
    35 mins ago






  • 1




    @DonAntonio editted. Thanks!
    – Ankit Kumar
    29 mins ago














8












8








8






The $n$th term of the sequence is $T_n=frac{n^2}{7^{n-1}}$.
To see that the series converges,
$$r=frac{a_{n+1}}{a_n}=frac{(n+1)^2cdot7^{n-1}}{7^ncdot n^2}$$
$$=frac{(1+1/n)^2}{7}$$
As n tends to $infty$, $r$ tends to $1/7$. Since $r<1$, the series converges. (Thanks to DonAntonio for this :) )



Note that
$$S=1+frac{4}{7}+frac{9}{7^2}+frac{16}{7^3}cdots$$
$$S-S/7=6S/7=1+frac{3}{7}+frac{5}{7^2}+frac{7}{7^3}cdots$$
$$6S/7-6S/49=1+frac{2}{7}+frac{2}{7^2}+frac{2}{7^3}cdots$$
$$frac{36S}{49}=1+frac{2}{7}cdotfrac{7}{6}=frac{4}{3}$$
$$implies S=frac{49}{27}$$






share|cite|improve this answer














The $n$th term of the sequence is $T_n=frac{n^2}{7^{n-1}}$.
To see that the series converges,
$$r=frac{a_{n+1}}{a_n}=frac{(n+1)^2cdot7^{n-1}}{7^ncdot n^2}$$
$$=frac{(1+1/n)^2}{7}$$
As n tends to $infty$, $r$ tends to $1/7$. Since $r<1$, the series converges. (Thanks to DonAntonio for this :) )



Note that
$$S=1+frac{4}{7}+frac{9}{7^2}+frac{16}{7^3}cdots$$
$$S-S/7=6S/7=1+frac{3}{7}+frac{5}{7^2}+frac{7}{7^3}cdots$$
$$6S/7-6S/49=1+frac{2}{7}+frac{2}{7^2}+frac{2}{7^3}cdots$$
$$frac{36S}{49}=1+frac{2}{7}cdotfrac{7}{6}=frac{4}{3}$$
$$implies S=frac{49}{27}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 29 mins ago

























answered 38 mins ago









Ankit Kumar

1,28017




1,28017








  • 2




    Nice...but it must first be established the first series converges
    – DonAntonio
    35 mins ago








  • 1




    Thanks for solution sir.
    – Ashish Yadav
    35 mins ago






  • 1




    @DonAntonio editted. Thanks!
    – Ankit Kumar
    29 mins ago














  • 2




    Nice...but it must first be established the first series converges
    – DonAntonio
    35 mins ago








  • 1




    Thanks for solution sir.
    – Ashish Yadav
    35 mins ago






  • 1




    @DonAntonio editted. Thanks!
    – Ankit Kumar
    29 mins ago








2




2




Nice...but it must first be established the first series converges
– DonAntonio
35 mins ago






Nice...but it must first be established the first series converges
– DonAntonio
35 mins ago






1




1




Thanks for solution sir.
– Ashish Yadav
35 mins ago




Thanks for solution sir.
– Ashish Yadav
35 mins ago




1




1




@DonAntonio editted. Thanks!
– Ankit Kumar
29 mins ago




@DonAntonio editted. Thanks!
– Ankit Kumar
29 mins ago










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