Evaluating $\lim_{n\to\infty} \left({1\over1\cdot2\cdot3}+{1\over2\cdot3\cdot4}+\cdots+{1\over(n-1)\cdot n\cdot(n+1)}\right)$

Question

The original question was to find $L=\displaystyle\lim_{n\to\infty}\sum_{k=1}^na_k$ where $a_n=\displaystyle{n\over(1+2+\cdots+(n-1))(1+2+\cdots+n)}$,

which I managed to get down to evaluating the following:

$$ L=4\lim_{n\to\infty} \left({1\over1\cdot2\cdot3}+{1\over2\cdot3\cdot4}+\cdots+{1\over(n-1)\cdot n\cdot(n+1)}\right) $$

I've tried looking around the internet (as much as you can with these things), and the only thing I was able to find is a method that uses definite integrals. I also found that the partial sum is $$\displaystyle {(n+2)(n+1)\over 4n(n+1)},$$ but this is definitely not something a student should have on the top of the head during the exam.

A hint would be greatly appreciated.

Answer 1

Expand by partial fractions:

$$\frac{1}{(n - 1)n(n + 1)} = -\frac 1 n + \frac 1 2 \frac 1 {n - 1} + \frac 1 2 \frac 1 {n + 1}$$

Now notice that a lot of the terms will cancel, and the series telescopes.

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This is similar to, and motivated by the fact that

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

telescopes to leave $1$.

Answer 2

We first note that $$ \frac{1}{(k-1)k(k+1)}=\frac{1}{2}\left(\frac{1}{k(k-1)}-\frac{1}{k(k+1)}\right), $$ and thus $$ 4\sum_{k=1}^{n-1} \frac{1}{k(k+1)(k+2)}=2\sum_{k=1}^{n-1}\left(\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)}\right)=\frac{2}{1\cdot 2}-\frac{2}{n(n+1)}. $$ Consequently $$ \lim_{n\to\infty}4\sum_{k=1}^{n-1} \frac{1}{k(k+1)(k+2)}=1. $$

Answer 3

More generally, $$\frac{1}{n(n+1)\dots(n+k)} = \frac{1}k\left(\frac{1}{n(n+1)\dots(n+k-1)}-\frac{1}{(n+1)\dots(n+k)}\right)$$ so $$\sum_{n=1}^\infty \frac{1}{n(n+1)\cdots (n+k)}$$ is a telescoping series with value $\frac{1}{k\cdot k!}$.

Answer 4

If you don't know how you should decompose a fraction, you can always use this method:

Let $$\frac{1}{(k-1)k(k+1)}=\frac{A}{k-1}+\frac{B}{k}+\frac{C}{k+1}$$

Then we have

$$1=Ak(k+1)+B(k-1)(k+1)+Ck(k-1)$$

$$\Leftrightarrow 1=Ak^2+Ak+Bk^2-B+Ck^2-Ck$$

$$\Leftrightarrow 1=(A+B+C)k^2+(A-C)k-B$$

Clearly, $A+B+C=0$, $A-C=0$ and $B=-1$

$\Rightarrow A=B=1/2$

Thus

$$\frac{1}{(k-1)k(k+1)}=\frac{1}{2(k-1)}-\frac{1}{k}+\frac{1}{2(k+1)}$$