Find lim as x to 0+ sum from n=1 to infinity

Find $\displaystyle\lim_{x\to0^+}\sum_{n=1}^{\infty}\frac1{(x+n)(x+n+1)}$.

Step 1 · Partial fractions

$$ \frac1{(x+n)(x+n+1)} = \frac1{x+n} - \frac1{x+n+1}. $$

Step 2 · Telescoping sum

The $N$-th partial sum is $$ S_N = \sum_{n=1}^{N}\left(\frac1{x+n}-\frac1{x+n+1}\right) = \frac1{x+1} - \frac1{x+N+1}. $$ For $x>0$, $\frac1{x+N+1}\to0$ as $N\to\infty$, so $$ \sum_{n=1}^{\infty}\frac1{(x+n)(x+n+1)} = \frac1{x+1}. $$

Step 3 · Take the limit

Since $\dfrac1{x+1}$ is continuous at $x=0$: $$ \lim_{x\to0^+}\sum_{n=1}^{\infty}\frac1{(x+n)(x+n+1)} = \lim_{x\to0^+}\frac1{x+1} = 1. $$

Summary Table

Additional Notes

• This is the same telescoping mechanism used in Problem 186's Cauchy-criterion proof for $\sum1/[(x+n-1)(x+n)]$ on $[0,a]$ — both problems exploit the exact partial-fraction cancellation to bypass any convergence-test machinery entirely.
• Because the closed form $1/(x+1)$ is defined and continuous at $x=0$ itself, no delicate limiting argument is needed: the series and its closed form agree exactly for every $x>0$, and simply evaluating the (continuous) right-hand side at $x=0$ gives the answer.
• At $x=0$ directly, the series becomes $\sum1/(n(n+1))=1$ (the classic telescoping sum), confirming the limit computed here without even needing the closed form.