On a calculus exam I took recently, there were two problems.
- Find the sum of the series defined by $\sum_{n=1}^{\infty}\frac{1}{(n(n+1))}$
- 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?
Best Answer
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}