Evaluate the sum S for the series, for a constant k in N
Evaluate the sum $S$ for the series, for a constant $k \in \mathbb{N}$: $$ \sum_{n=0}^{\infty} \frac{1}{(n+k)(n+k+1)}. $$
Explanation of Method
This is the same telescoping structure as $\sum \frac{1}{n(n+1)}$, but with every index shifted up by the constant $k$ and the sum starting at $n=0$ instead of $n=1$. Partial fractions again split the general term into a difference of consecutive reciprocals, $\frac1{n+k}-\frac1{n+k+1}$, which telescopes to leave only the very first term, $\frac1k$, once the vanishing tail is dropped in the limit.
- Decompose the general term with partial fractions.
- Sum the telescoping terms to get a closed form for $S_N$.
- Take the limit as $N \to \infty$.
Step 1 · Partial fraction decomposition
The general term $a_n = \dfrac{1}{(n+k)(n+k+1)}$ is decomposed as:
$$ \frac{1}{(n+k)(n+k+1)} = \frac{A}{n+k} + \frac{B}{n+k+1}. $$
Multiplying by $(n+k)(n+k+1)$ gives $1 = A(n+k+1) + B(n+k)$. Setting $n=-k$ gives $A=1$; setting $n=-k-1$ gives $B=-1$. Thus:
$$ a_n = \frac{1}{n+k} - \frac{1}{n+k+1}. $$
Step 2 · Determine the partial sum $S_N$
This is a telescoping series. The $N$-th partial sum, summing from $n=0$ to $N$, is:
$$ S_N = \sum_{n=0}^{N} \left( \frac{1}{n+k} - \frac{1}{n+k+1} \right). $$
Writing out the first and last few terms:
$$ S_N = \left(\frac1k-\frac1{k+1}\right) + \left(\frac1{k+1}-\frac1{k+2}\right) + \cdots + \left(\frac1{N+k}-\frac1{N+k+1}\right). $$
All intermediate terms cancel, leaving only the first positive term and the last negative term:
$$ S_N = \frac{1}{k} - \frac{1}{N+k+1}. $$
Step 3 · Calculate the sum $S$
The sum $S$ is the limit of $S_N$ as $N \to \infty$:
$$ S = \lim_{N \to \infty} S_N = \lim_{N \to \infty} \left(\frac{1}{k} - \frac{1}{N+k+1}\right). $$
Since $k$ is fixed and $\lim_{N \to \infty} \dfrac{1}{N+k+1} = 0$:
$$ S = \frac{1}{k} - 0 = \frac{1}{k}. $$
Summary Table
| Quantity | Closed form |
|---|---|
| General term $a_n$ | $\dfrac{1}{n+k} - \dfrac{1}{n+k+1}$ |
| Partial sum $S_N$ | $\dfrac1k - \dfrac{1}{N+k+1}$ |
| Sum $S$ | $\dfrac1k$ |
Additional Notes
- Setting $k=1$ recovers Problem 3's series $\sum_{n=1}^\infty \frac{1}{n(n+1)}$ exactly (after reindexing $n \to n+1$), and the sum correctly reduces to $\dfrac11 = 1$.
- The sum $S=\dfrac1k$ decreases as $k$ grows, which makes sense: larger $k$ means every term in the series is smaller, since the series effectively starts further out along $\sum \frac{1}{n(n+1)}$.
- Because $k \in \mathbb{N}$, division by $k$ is always well-defined; if $k$ could be $0$ the first term $\frac{1}{k}$ would be undefined, so the natural-number restriction is essential.
Result
The partial sum is:
The sum of the series is: