$$\sum_{m=5}^\infty \frac{3}{m^2+3m+2}$$

Given this problem my first approach was to take the limit of partial sums. To my surprise this didn't work. Many expletives later I realized it was a telescoping series.

My question is why my first approach failed. My expectation is that both approaches would produce the same answer. Why didn't they?

First approach:

$$\sum_{m=5}^\infty \frac{3}{m^2+3m+2} = \lim \limits_{N \to \infty} \int_{m=5}^N \frac{3}{m^2+3m+2} = \lim \limits_{N \to \infty} 3 \left [ \ln \left ( \frac{N+1}{N+2} \right ) + \ln \left ( \frac{7}{6} \right ) \right ]$$$$ = 3 \ln(7/6) \approx 0.46245$$

An empirical check showed that the above approach is wrong. After I realized it was a telescoping series I was able to produce a sequence of partial sums:

$$S_{m} = \left ( \frac{3}{6}-\frac{3}{m+2} \right ) $$

And the limit of this sequence gets to an answer that agrees with a crude empirical spreadsheet validation:

$$\lim \limits_{m \to \infty} \left ( \frac{3}{6}-\frac{3}{m+2} \right ) = \frac{1}{2}$$

So clearly my initial intuition and understanding was wrong. But why was it wrong? I thought I could take the limit of an integral to calculate the value of a series. In what case do we apply the first approach I took? I've used it before, but I must have forgotten how to apply it correctly (and all my searches come up with references to computing convergence, not actual values).


The integral test simply tells you if an infinite sum is convergent, it will not necessarily tell you what the sum converges to. Imagine back to when you first started learning about the Riemann integral. You might remember seeing pictures of a smooth curve with rectangular bars pasted over the curve, approximating the area below the curve. The idea being that the more narrow the width you make the bars, the closer you will get to the actual area under the curve. The integral you calculated gives you the area under the curve as if $\frac{3}{m^2+3m+2}$ were a smooth curve. Here we think of $m$ taking on every real number from $5$ to $\infty$. The infinite sum gives you the area of the rectangles with height $\frac{3}{m^2+3m+2}$ and width $1$ (not at all narrow for a rectangle). In this case we strictly treat $m$ as an integer. As such, the integral and the sum can only be rough approximations of each other.

  • $\begingroup$ Ahh, of course, the sum isn't represented as a smooth curve. When you put it in those terms, and in terms of a Riemann sum of width 1, it makes sense. This limit of the integral is just a convergence test approach, which nicely explains why all my searches seemed to come up with convergence tests. To find the actual solution I need to find a sequence of partial sums (as I did after putting a skull shaped dent in my wall). $\endgroup$ – David Parks May 22 '15 at 4:28
  • $\begingroup$ @DavidParks you got it, could not have said it better :) $\endgroup$ – graydad May 22 '15 at 4:35

You can sometimes use this idea to show a series converges (or diverges) but it will not give you the sum. Think about it: an impoper integral is the limit of a Riemann sum over an interval $[a,b]$ on a continuous variable $x$, after which you take a limit $b\rightarrow \infty$. Your "integral" is over a discrete variable (the integers) and makes no sense. You actually found



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