# Convergence of $\sum_{n=1}^\infty 1/(n - c)^2$

I'm curious as to whether or not the following series converges,

$$\sum_{n=1}^\infty \frac{1}{(n - c)^2},$$

where $c$ is some positive constant, $c \notin \mathbb{Z}_{>0}$.

Initially my intuition lead me to believe that it did converge since, for large enough $n$, one could argue that it looks a lot like the series $\sum \frac{1}{n^2}$ , but so far i've been unable to throw together a formal proof. Any ideas or a proof of convergence/divergence would be much appreciated.

• Are we given that $c$ is less than $1$ or not an integer? Commented Oct 28, 2013 at 2:51
• If $n\neq c$ then you can make comparison with the series $\sum \frac{1}{n^2}$. Commented Oct 28, 2013 at 2:57
• Yea you guys are right. I was looking at the cases when $c \notin \mathbb{Z}_{>0}$. I'll throw that into the statement of the question. Commented Oct 28, 2013 at 3:22

Assuming that $c$ is a non-integer constant, we have the following argument:

There is some $n_0$ so that $n_0 - c>0$. With that, we may note that

$$\sum_{n=1}^\infty \frac{1}{(n-c)^2} = \sum_{n=1}^{n_0} \frac{1}{(n-c)^2} + \sum_{k=1}^\infty \frac{1}{(k+(n_0-c))^2}$$ Now, consider that second sum. We note that $$\frac{1}{(k+(n_0-c))^2} < \frac{1}{k^2}$$ Since the sum $\sum_{k=1}^\infty \frac 1{k^2}$ converges, we may state by the comparison test that $$\sum_{k=1}^\infty \frac{1}{(k+(n_0-c))^2}$$ converges, which means that our original sum must converge.

• Ah yes. That's really nice, I wasn't quite sure how to formalize that argument. Thanks a lot! Commented Oct 28, 2013 at 3:20
• Should probably be a $k^2$ in the third to last line. Commented Oct 28, 2013 at 4:11
• @AndrewRoss you're welcome! Commented Oct 28, 2013 at 13:22
• @Bitrex thanks, duly noted. Commented Oct 28, 2013 at 13:22

use the integral test:

$$\int\limits_{[1, \infty]} \frac{dx}{(x-c)^2} = \int\limits_{[1,\infty]} \frac{d(x-c)}{(x-c)^2} = \lim_{ t \to \infty} \frac{-1}{(t-c)^2} + \frac{1}{(1-c)^2} = \frac{1}{(1-c)^2}$$

Therefore, your series must converge by integral test.

• I hadn't thought about approaching the question like this at all, but I really like this argument as well. Commented Oct 28, 2013 at 3:25