Evaluate the sum S for the series: S = sum from n=1
Evaluate the sum $S$ for the series: $$ S = \sum_{n=1}^{\infty}\left[\frac{1}{n^2}-\frac{1}{(n+1)^k}\right]. $$
Explanation of Method
Rather than a telescoping cancellation, this series splits cleanly into two separate $p$-series: one with fixed exponent $2$, and one with exponent $k$ shifted by one index. Each piece is a known object — a Basel-problem sum and a shifted Riemann zeta series — so the strategy is to determine when each piece converges, evaluate each in terms of $\zeta$, and subtract.
- Split the series into two $p$-series and determine when each converges.
- Express the shifted series in terms of the Riemann zeta function $\zeta(k)$.
- Combine the two pieces into a single closed-form sum.
Step 1 · Condition for convergence
Split the series into two components:
$$ S = \sum_{n=1}^{\infty} \frac{1}{n^2} - \sum_{n=1}^{\infty} \frac{1}{(n+1)^k} = S_1 - S_2. $$
A series $\sum a_n$ may be split into $\sum b_n - \sum c_n$ only if both $\sum b_n$ and $\sum c_n$ converge.
- $S_1 = \sum_{n=1}^{\infty} \dfrac{1}{n^2}$ is a convergent $p$-series ($p=2>1$), equal to the Basel problem sum $\zeta(2) = \dfrac{\pi^2}{6}$.
- $S_2 = \sum_{n=1}^{\infty} \dfrac{1}{(n+1)^k}$ is a shifted $p$-series, convergent if and only if $k>1$.
Therefore the total series $S$ converges if and only if $k > 1$.
Step 2 · Calculating the sum $S_2$
For $k>1$, $S_2$ relates directly to $\zeta(k) = \sum_{n=1}^{\infty}\frac1{n^k}$:
$$ S_2 = \sum_{n=1}^{\infty} \frac{1}{(n+1)^k} = \frac{1}{2^k}+\frac{1}{3^k}+\frac1{4^k}+\cdots $$
This is $\zeta(k)$ with the $n=1$ term removed:
$$ S_2 = \left(\sum_{n=1}^\infty \frac1{n^k}\right) - \frac{1}{1^k} = \zeta(k) - 1. $$
Step 3 · Calculating the total sum $S$
For $k > 1$:
$$ S = S_1 - S_2 = \zeta(2) - (\zeta(k)-1) = \frac{\pi^2}{6} - \zeta(k) + 1. $$
Special case $k=2$: the series becomes telescoping, $\sum_{n=1}^{\infty}\left[\frac1{n^2}-\frac1{(n+1)^2}\right]$, with partial sum $S_N = 1-\frac1{(N+1)^2} \to 1$. This matches the general formula:
$$ S = \frac{\pi^2}{6}-\zeta(2)+1 = \frac{\pi^2}{6}-\frac{\pi^2}{6}+1 = 1. $$
Summary Table
| Component | Value |
|---|---|
| $S_1 = \sum \frac1{n^2}$ | $\zeta(2) = \dfrac{\pi^2}{6}$ |
| $S_2 = \sum \frac1{(n+1)^k}$ | $\zeta(k) - 1$, requires $k>1$ |
| Convergence condition | $k > 1$ |
| Sum $S$ | $1 + \dfrac{\pi^2}{6} - \zeta(k)$ |
Additional Notes
- For $1 < k < 2$ (e.g. $k=1.5$), $S_2$ still converges but $\zeta(k) > \zeta(2)$, so $S = 1+\frac{\pi^2}{6}-\zeta(k)$ can be smaller than $1$, or even negative for $k$ close to $1$, since $\zeta(k)\to\infty$ as $k\to1^+$.
- For $k \le 1$, $S_2$ diverges (harmonic-like behavior), so the two-piece split is invalid and $S$ has no finite value — the original series itself diverges in that regime.
- This problem illustrates that not every series needing "$S_1 - S_2$" splitting is telescoping; here the two pieces are genuinely independent $p$-series, unlike Problems 3–5, and the shared index shift is what connects $S_2$ to $\zeta(k)$.
Result
The series converges if and only if $k > 1$. In that case, the sum is: