On a calculus exam I took recently, there were two problems.

  1. Find the sum of the series defined by $\sum_{n=1}^{\infty}\frac{1}{(n(n+1))}$
  2. A series $\sum_{n=1}^{\infty}a_n$ has partial sums $s_n = \frac{n-1}{n+1}$. Find $a_n$ and $\sum_{n=1}^{\infty}a_n$

I got both problems correct, but my answers seem to conflict with each other.

For 1, I got 1.

For 2, I also got 1 for the sum, and $\frac{2}{n(n+1)}$ for the value of $a_n$

My question is, how can this be possible? The value of $a_n$ in question 2 should be twice that of question 1, so shouldn't the value of the series in question 2 be 2? But when you look at the definition of the series in question 2, the sum as n approaches infinity is obviously 1.

Did I make a mistake in finding $a_n$? I got it from $s_n - s_{n-1}$, which should just give me $a_n$. I brought it up with my TA and professor, and they didn't know, either. What am I missing here?


2 Answers 2


In 2, $a_n=s_n-s_{n-1}=\frac{2}{n(n+1)}$ for $n>1$, but $a_1=s_1=0$, since $s_0$ is not defined. Therefore, \begin{align} \sum_{n=1}^{\infty}a_n&=0+\sum_{n=2}^{\infty}\frac{2}{n(n+1)} \\ &=2\sum_{n=2}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right) \\ &=2\cdot\frac{1}{2}=1. \end{align}

  • $\begingroup$ Can you explain why the $a_n$ found is only valid for n > 1? I don't understand why the start point would change. I can see that it doesn't apply for n = 1, but am not sure why. $\endgroup$
    – Suop
    Nov 21 at 5:26
  • 1
    $\begingroup$ Notice that $s_n=\sum_{k=1}^n a_k$, so it is defined only for $n\geq 1$. In particular, $s_1=a_1$. $\endgroup$
    – Gonçalo
    Nov 21 at 5:32
  • 2
    $\begingroup$ Here is a perhaps simpler version of this kind of mistake. $\endgroup$
    – Dan
    Nov 21 at 21:35
  • 2
    $\begingroup$ Subtle. So if a series starts from $n=1$, you reconstruct the terms $a_n$ from the partial sums $s_n$ by saying $a_n=s_n - s_{n-1}$ for $n>1$ but the initial term is $a_1=s_1$. Of course, if you use the natural convention $s_0=0$ (empty sum), then $a_n=s_n - s_{n-1}$ holds for $n\ge 1$. However, in problem 2, the formula $s_n = \frac{n-1}{n+1}$ given does not automatically produce zero if you plug in $n=0$ into it. It would give $s_0=-1$ which is absurd. So you have to assume the formula in the problem is for $n>0$. $\endgroup$ Nov 22 at 9:35
  • $\begingroup$ Good answer! The takeaway from this for me is, two quite different series can have deceptively similar terms in their sequences. “How it changes” is only half the story; we also need “where it starts”. (Which is also the lesson when encountering the constant of integration…) $\endgroup$ Nov 22 at 14:00

Gonçalo’s answer is a very elegant look at the matter. But I think there is another approach that may make it easier to grasp why this is the result: look at some terms of the sequence, and some partial sums.

In question 1, substituting $n = 1, 2, 3, 4, 5, …$ into $a_n = \frac{1}{n \left(n + 1 \right)}$, we get:

$$a_n = \frac{1}{2}, \frac{1}{6}, \frac{1}{12}, \frac{1}{20}, \frac{1}{30}, …$$

Summing these gives us:

$$s_n = \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, …$$

I’ll leave a rigorous proof as an exercise for the reader 😉︎, but it’s evident that here, $s_n = \frac{n}{n + 1}$.

In question 2, we work in the opposite order. Substituting $n = 1, 2, 3, 4, 5, …$ into $s_n = \frac{n - 1}{n + 1}$, we get:

$$s_n = 0, \frac{1}{3}, \frac{1}{2}, \frac{3}{5}, \frac{2}{3}, …$$

Taking the difference gets us the terms being summed:

$$a_n = 0, \frac{1}{3}, \frac{1}{6}, \frac{1}{10}, \frac{5}{6}, …$$

As you observed, this is a sequence where $a_n = \frac{2}{n \left(n + 1 \right)}$. But as Gonçalo pointed out, that doesn’t hold for $n = 1$.

If it did, then indeed $\sum_{n=1}^{\infty} \frac{2}{n \left(n + 1 \right)} = 2$, twice the value of $\sum_{n=1}^{\infty} \frac{1}{n \left(n + 1 \right)}$ in question 1. But that sequence would have a first term of $a_1 = 1$ (and partial sums $s_n = \frac{2n}{n + 1}$). In what we’re actually given, the first term is lower by 1, and hence so is the value of the series.


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