1

$$\sum_{n=1}^{\infty}\frac{14}{49n^2-84n-13}$$ my steps were:

$\sum_{n=1}^{\infty}\frac{14}{49n^2-84n-13}$=$\sum_{n=1}^{\infty}\frac{14}{(7n+1)(7n-13)}$=$\sum_{n=1}^{\infty}\frac{1}{7n-13}-\frac{1}{7n+1}$

I was trying count if n=1,n=2,n=3,n=4 and after find the whole sum of this

And my answer is: $$\frac{721}{957}$$ Did I make a mistake somewhere? Thank you for helping in advance!

Akane123
  • 57
  • 1
    If the result is a rational , it probably involved telescoping. – mick Nov 29 '23 at 22:42
  • 1
    @mick is right. The series is of the form $\sum_{n=1}^\infty(a_n-a_{n+2})=a_1+a_2$. What you've computed is $\sum_{n=1}^4(a_n-a_{n+2})=a_1+a_2-a_5-a_6$. – J.G. Nov 29 '23 at 22:44

2 Answers2

3

Note that$$\sum_{n=1}^\infty\left(\frac1{7n-13}-\frac1{7n+1}\right)=\sum_{n=1}^\infty\left(\left(\frac1{7n-13}-\frac1{7n-6}\right)+\left(\frac1{7n-6}-\frac1{7n+1}\right)\right).$$But both series$$\sum_{n=1}^\infty\left(\frac1{7n-13}-\frac1{7n-6}\right)\quad\text{and}\quad\sum_{n=1}^\infty\left(\frac1{7n-6}-\frac1{7n+1}\right)$$are telescoping series; their sums are $-\frac16$ and $1$ respectively.

Therefore, the sum of your series is $\frac56$.

José Carlos Santos
  • 427,504
1

We first consider its partial sum which is telescoping. $$ \begin{aligned} S_N & =\sum_{n=1}^N\left(\frac{1}{7 n-13}-\frac{1}{7 n+1}\right) \\ & =\sum_{n=1}^N \frac{1}{7(n-2)+1}-\sum_{n=1}^N \frac{1}{7 n+1} \\ & =-\frac{1}{6}+1+\sum_{n=3}^N \frac{1}{7(n-2)+1}-\sum_{n=1}^N \frac{1}{7 n+1} \\ & =\frac{5}{6}+\sum_{n=1}^{N-2} \frac{1}{7 n+1}-\sum_{n=1}^N \frac{1}{7 n+1} \\ & =\frac{5}{6}-\frac{1}{7(N-1)+1}-\frac{1}{7 N+1} \end{aligned} $$ As $N \rightarrow \infty$, $$ \begin{aligned} S_{\infty} & =\lim _{N \rightarrow \infty} \sum_{n=1}^N \frac{14}{49n^2-84 n-13} \\ & =\lim _{N \rightarrow \infty}\left(\frac{5}{6}-\frac{1}{7 N-6}-\frac{1}{7 N+1}\right) \\ & =\frac{5}{6} \end{aligned} $$

Lai
  • 20,421