Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Please help me to find the values of $x$ in which the series $$\sum_{n=1}^{\infty}\frac{1}{(x+n)(x+n-1)}$$ converges? I applied some tests for it but...:(

Thank you

share|improve this question
@BrettFrankel: I fixed the summation. –  Nancy Rutkowskie Dec 25 '12 at 13:55
add comment

4 Answers

up vote 4 down vote accepted



share|improve this answer
add comment

We can use partial fractions to decompose this fraction. We need to find $A$ and $B$ such that

$$\frac{A}{x+n} + \frac{B}{x+n-1} \equiv \frac{1}{(x+n)(x+n-1)} \, . $$

If we cross multiply, we get $A(x+n-1)+B(x+n)\equiv1$. When $x=-n$ we get $-A\equiv1$ and when $x=1-n$ we get $B \equiv 1$. It now follows that:

$$\frac{1}{(x+n)(x+n-1)} \equiv \frac{1}{x+n-1} - \frac{1}{x+n} \, . $$

Let's examine these terms as $n$ advances from $1$. Substituting $n=1,2,3,\ldots$ gives:

$$\left( \frac{1}{x} - \frac{1}{x+1} \right) + \left( \frac{1}{x+1} - \frac{1}{x+2} \right) + \left( \frac{1}{x+2} - \frac{1}{x+3} \right) + \cdots = $$

$$\frac{1}{x} - \frac{1}{x+1} + \frac{1}{x+1} - \frac{1}{x+2} + \frac{1}{x+2} - \frac{1}{x+3} + \cdots $$

You should be able to see that each term is cancelled by the very next term. For this to be well-defined we need the denominators to be non-zero and for the terms to tend to zero. With that in mind, $x$ can be anything but zero or a negative integer. With that in mind we have:

$$\sum_{n=1}^{\infty} \frac{1}{(x+n)(x+n-1)} = \frac{1}{x} \, . $$

share|improve this answer
add comment

Besides to @N.S. nice hit. Think about $S_n$ and find the values that the following limit exists: $$\lim_{n\to\infty}S_n=1/x$$

share|improve this answer
+++++++++++++++++++++++ –  amWhy Mar 3 '13 at 0:11
add comment

The approach given by N.S. and Babak here is elegant. Here's something a little more direct, which doesn't require you to be clever.

First, figure out which values of $x$ would give you a term with zero in the denominator. Those are obviously bad values for $x$.

Next, note that all but finitely many terms will be positive (in fact all terms will be positive if $x$ is positive), so if the series diverges, the sum must be $+\infty$.

Let's assume $x>1$. If $x<1$, we can let $y$ be $x$ plus some large integer (so that $y>1$) and write $$\sum_{n=1}^\infty\frac{1}{(x+n)(x+n-1)}=\sum_{n=1}^\infty\frac{1}{(y+n)(y+n-1)}+\text{finitely many terms}$$

Now $$\sum_{n=1}^\infty\frac{1}{(x+n)(x+n-1)}<\sum_{n=1}^\infty\frac{1}{(x+n-1)(x+n-1)}=\sum_{n=1}^\infty\frac{1}{(x+n-1)^2}<\sum_{n=1}^\infty\frac{1}{n^2}$$

That last sum is a convergent $p-$series.

share|improve this answer
I think there might be a mistake here. Since $x+n - 1 < x+n$ it follows that $1/(x+n-1) > 1/(x+n)$ and so the first inequality is not valid. As a counter example, let $x=0.5$. The series converges to $2$ but $2 > \pi^2/6$. –  Fly by Night Dec 25 '12 at 15:11
There's an off-by-one error in the last step, but the core argument is correct. –  Steven Stadnicki Dec 25 '12 at 16:34
@StevenStadnicki Could you please address the counter example? Putting $x = 0.5$, the argument seems to claim that: $$\sum_{n=1}^{\infty}\frac{4}{4n^2-1} < \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} $$ Besides that, the OP asked for the values of $x$ for which the sum converges. while, in reality: $$\sum_{n=1}^{\infty}\frac{4}{4n^2-1} = 2 $$ and $2$ is not less than $\pi^2/6 \approx 1.64$. Besides that, the OP asked for the values of $x$ for which the sum converges. –  Fly by Night Dec 25 '12 at 21:10
@FlybyNight It's true that $x+n-1\lt x+n$, but while that means that $\dfrac{1}{(x+n)(x+n+1)}\gt\dfrac{1}{(x+n)^2}$, it also means that $\dfrac{1}{(x+n)(x+n-1)}\lt\dfrac{1}{(x+n-1)^2}$, so by 'bumping' the indices by one the convergence (which is all the OP cares about) still holds. –  Steven Stadnicki Dec 26 '12 at 3:46
@StevenStadnicki Please address my counter example. Besides that, the OP asks about the values of $x$ for which the series converges. That question hasn't been addressed. –  Fly by Night Dec 26 '12 at 20:38
show 3 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.