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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

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@BrettFrankel: I fixed the summation. –  Nancy Rutkowskie Dec 25 '12 at 13:55
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4 Answers

up vote 4 down vote accepted

Hint:

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

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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:

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

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

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+++++++++++++++++++++++ –  amWhy Mar 3 '13 at 0:11
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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.

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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
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