Evaluate the sum S for the series: sum from n=1
Evaluate the sum $S$ for the series: $$ \sum_{n=1}^{\infty} \frac{2n+1}{n^2(n+1)^2}. $$
Explanation of Method
The numerator $2n+1$ is exactly the difference $(n+1)^2 - n^2$, which is the key observation that unlocks this problem. Splitting the fraction along that difference converts the general term into $\frac1{n^2} - \frac1{(n+1)^2}$ — a difference of consecutive terms of the sequence $b_n = \frac1{n^2}$ — so the series telescopes just as the reciprocal-product series do, but with squares instead of simple reciprocals.
- Rewrite the numerator as a difference of consecutive squares.
- Split the general term into $\dfrac1{n^2} - \dfrac1{(n+1)^2}$.
- Sum the telescoping terms and take the limit.
Step 1 · Decompose the general term
The general term is $a_n = \dfrac{2n+1}{n^2(n+1)^2}$. Notice that the numerator equals a difference of squares:
$$ (n+1)^2 - n^2 = (n^2+2n+1) - n^2 = 2n+1. $$
Substituting this into $a_n$ and splitting the fraction:
$$ a_n = \frac{(n+1)^2 - n^2}{n^2(n+1)^2} = \frac{(n+1)^2}{n^2(n+1)^2} - \frac{n^2}{n^2(n+1)^2} = \frac{1}{n^2} - \frac{1}{(n+1)^2}. $$
This decomposes $a_n$ into the required telescoping form $a_n = b_n - b_{n+1}$, with $b_n = \dfrac1{n^2}$.
Step 2 · Determine the partial sum $S_N$
The $N$-th partial sum is:
$$ S_N = \sum_{n=1}^{N} \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right). $$
Writing out the terms shows the telescoping cancellation:
$$ S_N = \left(\frac1{1^2}-\frac1{2^2}\right) + \left(\frac1{2^2}-\frac1{3^2}\right) + \cdots + \left(\frac1{(N-1)^2}-\frac1{N^2}\right) + \left(\frac1{N^2}-\frac1{(N+1)^2}\right). $$
After cancellation, only the first and last pieces survive:
$$ S_N = \frac{1}{1^2} - \frac{1}{(N+1)^2} = 1 - \frac{1}{(N+1)^2}. $$
Step 3 · Calculate the sum $S$
The sum $S$ is the limit of $S_N$ as $N \to \infty$:
$$ S = \lim_{N \to \infty} S_N = \lim_{N \to \infty} \left(1 - \frac{1}{(N+1)^2}\right). $$
Since $\lim_{N \to \infty} \dfrac{1}{(N+1)^2} = 0$:
$$ S = 1 - 0 = 1. $$
Summary Table
| Quantity | Closed form |
|---|---|
| General term $a_n$ | $\dfrac1{n^2} - \dfrac1{(n+1)^2}$ |
| Partial sum $S_N$ | $1 - \dfrac{1}{(N+1)^2}$ |
| Sum $S$ | $1$ |
Additional Notes
- The trick of writing the numerator as a difference of squares, $(n+1)^2-n^2$, is what reveals the telescoping structure; without it, the fraction $\frac{2n+1}{n^2(n+1)^2}$ gives no obvious hint of cancellation.
- Coincidentally the sum equals $1$, the same value as Problem 3's series $\sum \frac{1}{n(n+1)}$ — but the two series are different (this one decays like $n^{-3}$ for large $n$, versus $n^{-2}$ there), so the matching sums are not related by any simple identity.
- Since every term $a_n > 0$, the partial sums $S_N = 1 - \frac{1}{(N+1)^2}$ increase monotonically and stay strictly below $1$, approaching it only in the limit.
Result
The partial sum is:
The sum of the series is: