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
    1 hour ago










  • I'm sorry. I wrote it wrong.
    – Ashish Yadav
    1 hour 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
    1 hour ago










  • I'm sorry. I wrote it wrong.
    – Ashish Yadav
    1 hour 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 1 hour 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 1 hour 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
    1 hour ago










  • I'm sorry. I wrote it wrong.
    – Ashish Yadav
    1 hour ago














  • 5




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










  • I'm sorry. I wrote it wrong.
    – Ashish Yadav
    1 hour ago








5




5




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




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












I'm sorry. I wrote it wrong.
– Ashish Yadav
1 hour ago




I'm sorry. I wrote it wrong.
– Ashish Yadav
1 hour 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
    59 mins ago








  • 1




    Thanks for solution sir.
    – Ashish Yadav
    59 mins ago






  • 1




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











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1 Answer
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1 Answer
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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
    59 mins ago








  • 1




    Thanks for solution sir.
    – Ashish Yadav
    59 mins ago






  • 1




    @DonAntonio editted. Thanks!
    – Ankit Kumar
    53 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
    59 mins ago








  • 1




    Thanks for solution sir.
    – Ashish Yadav
    59 mins ago






  • 1




    @DonAntonio editted. Thanks!
    – Ankit Kumar
    53 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 54 mins ago

























answered 1 hour ago









Ankit Kumar

1,28717




1,28717








  • 2




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








  • 1




    Thanks for solution sir.
    – Ashish Yadav
    59 mins ago






  • 1




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














  • 2




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








  • 1




    Thanks for solution sir.
    – Ashish Yadav
    59 mins ago






  • 1




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








2




2




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






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






1




1




Thanks for solution sir.
– Ashish Yadav
59 mins ago




Thanks for solution sir.
– Ashish Yadav
59 mins ago




1




1




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




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










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