Sum of a Series

Though others have pointed out that the series telescopes, you may be interested to know that we can also find the sum by explicitly calculating the partial sums and taking the limit. To do this, we'll use the method of finite calculus (see: Finite Calculus: A Tutorial for Solving Nasty Sums [PDF]). Along the way we'll make use of falling powers--sometimes called falling factorials--which are central to finite calculus (the PDF linked above contains a great explanation of them).

Here be LaTeX. Check the sidebar a la derecha for the addon to display it.

We want to calculate the k^th partial sum of the series given. We'll pull out the 4 so that the series in question is just

    [; \sum_{n = 2}^{\infty} \frac{1}{n(n+2)}. ;]

Once we evaluate that, we'll just multiply the result by 4 to get the answer to the original question. Now, the partial sums we want to calculate are

    [; S_k = \sum_{n = 2}^{k} \frac{1}{n(n+2)}. ;]

By the fundamental theorem of finite calculus, we can rewrite this sum as a "definite integral-like" sum. Notice that we increase the upper limit of the sum by 1.

    [; \sum_{n = 2}^{k} \frac{1}{n(n+2)} = \sum_{2}^{k+1} \frac{1}{n(n+2)} \delta n. ;]

The game plan is to rewrite the summand to involve falling powers (which are indicated with an underline), then apply the discrete integration by parts formula to find an explicit formula for the partial sums [; S_k ;]. So, first, a little manipulation.

    [; \sum_{2}^{k+1} \frac{1}{n(n+2)} \delta n ;]
            [; = \sum_{2}^{k+1} \frac{n+1}{n(n+1)(n+2)} \delta n ;]
            [; = \sum_{2}^{k+1} (n+1)(n-1)^{\underline{-3}} \delta n. ;]

Next, we use the power rule for falling powers to calculate the discrete derivative

    [; \Delta[(-1/2)(n-1)^{\underline{-2}}] = (n-1)^{\underline{-3}}, ;]

so that we can rewrite the sum as

    [; \sum_{2}^{k+1} (n+1) \Delta [(-1/2)(n-1)^{\underline{-2}}] \delta n. ;]

This resembles the form for integrals ripe for integration by parts, no? Indeed, we can apply discrete integration by parts to transform the sum into

    [; -\frac{1}{2} \left[(n+1)(n-1)^{\underline{-2}}\right]_{2}^{k+1} + \frac{1}{2} \sum_{2}^{k+1} n^{\underline{-2}} \delta n. ;]

Finally, the remaining integral is a simple application of the power rule for falling powers, so we have

    [; -\frac{1}{2} \left[(n+1)(n-1)^{\underline{-2}}\right]_{2}^{k+1} - \frac{1}{2}\left[n^{\underline{-1}}\right]_{2}^{k+1}. ;]

Before plugging in the limits let's clean things up a little by transforming the falling powers back into the usual notation.

    [; -\frac{1}{2} \left[\frac{1}{n}\right]_{2}^{k+1} - \frac{1}{2}\left[\frac{1}{n+1}\right]_{2}^{k+1}. ;]

There, that's a little nicer. Now, just plug in the limits like a regular calculus integral. We end up with an explicit form for the partial sum.

    [; S_k = \frac{5}{12} - \frac{2(k+1)+1}{2(k+1)(k+2)}. ;]

Remember that we factored out a 4 at the beginning, so multiplying this expression by 4 gives us the partial sums of the original series.

    [; \sum_{n = 2}^{k} \frac{4}{n(n+2)} = \frac{5}{3} - \frac{4(k+1)+2}{(k+1)(k+2)}. ;]

The denominator of the term on the right is quadratic in k while the numerator is linear, so we see that the limit of the partial sums is 5/3. In other words,

    [; \sum_{n = 2}^{\infty} \frac{1}{n(n+2)} = \lim_{k \to \infty} \sum_{n = 2}^{k} \frac{4}{n(n+2)} = \frac{5}{3}. ;]

Nifty, huh? This method gives us more information than just the limit of the series; in many cases, finding the partial sums is much more useful.

Partial fractions + telescopic cancellation => 5/3, I think.

Try doing a partial fraction decomposition, write out a few partial sums, and see if a pattern emerges.

http://www.wolframalpha.com/input/?i=sum+of+4%2F%28n%28n%2B2%29%29+from+n+%3D+2+to+infinity

It looks like it's probably going to be a telescoping series.

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