Let $S=\sum\limits_{n=1}^\infty a_n$ be an infinite series such that $S_N=4-\frac{2}{N^2}$.

(a) Find a general formula for $a_n$.

(b) Find the sum $\sum\limits_{n=1}^\infty a_n$.

Can you explain to me how I can convert the partial sum to the general equation?

(a) What are the values of $\sum\limits_{n=1}^{10} a_n$ and $\sum\limits_{n=4}^{16} a_n$?

$\sum\limits_{n=1}^{10} a_n=23433271/635040$

$\sum\limits_{n=4}^{16} a_n= 15799025474051/259718659200 $

(b) What is the value of $a_3$? $167/18$

Why aren't these values correct as well?


Here are some hints.

(a) You make $S_n$ by adding $a_n$ to $S_{n-1}$.

(b) What is the definition of $\sum_{n=1}^\infty a_n$?

  • $\begingroup$ I know that it's something-2/n^2 and I know that something is equal to -2/(N-1)^2 which is also equal to 4 $\endgroup$ – user96246 Apr 28 '14 at 2:45
  • $\begingroup$ I don't understand. What is "it" in your comment? $\endgroup$ – user134824 Apr 28 '14 at 2:46
  • $\begingroup$ the value of SN-1 $\endgroup$ – user96246 Apr 28 '14 at 2:47
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    $\begingroup$ Ahhh I think I just realized what SN is. It's the N'th term of the partial sum right? $\endgroup$ – user96246 Apr 28 '14 at 3:05
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    $\begingroup$ Yes! Sorry if that wasn't clear. This problem would be impossible if you didn't know it. $\endgroup$ – user134824 Apr 28 '14 at 3:06

Note that $S_{N}-S_{N-1}=a_N$, so that shall give you $a_n$

  • $\begingroup$ I don't get how that's it... $\endgroup$ – user96246 Apr 28 '14 at 23:38
  • $\begingroup$ @user96246: use definition of $S_N$. $S_N=a_1+a_2+..a_N$. So, $S_{N+1}-S_N=a_1+a_2+..a_N+a_{N+1}-(a_1+a_2+..a_N)=a_{N+1}$ $\endgroup$ – voldemort Apr 29 '14 at 0:38
  • $\begingroup$ ok but how would you find SN+1? $\endgroup$ – user96246 Apr 29 '14 at 0:39
  • $\begingroup$ @user96246: You have a formula for $S_n$, right? Put $n=N+1$ in yor formula. $\endgroup$ – voldemort Apr 29 '14 at 0:41
  • $\begingroup$ But wouldn't that defy the limit? by limit i mean N approaching infinity? $\endgroup$ – user96246 Apr 29 '14 at 0:48

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