# Let $x_{n}>0$ be such that $\sum_{2}^{\infty}x_{n}$ diverges

I came across the following problem that says:

Let $x_{n}>0$ be such that $\sum_{2}^{\infty}x_{n}$ diverges and $\sum_{2}^{\infty}x_{n}^2$ converges. Then $x_n$ can not be which of the following?
\begin{align}\frac{n}{n^{2}+1} \tag{a}\\ \frac{\log n}{n}\tag{b}\\ \frac{1}{n \sqrt {\log n}}\tag{c}\\\frac{1}{n(\log n)^{2}}\tag{d}\end{align}

Can someone point me in the right direction? Thanks in advance for your time.

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Hint: The answer is (d). The series $\sum\frac{1}{n\log^2 n}$ converges. To show this, consider instead the integral $\int \frac{dx}{x\log^2 x}$. –  Anon Jan 10 '13 at 10:35
There is a problem with this question. You need the series to start at $n = 2$, otherwise dividing by $\log{n}$ includes dividing by zero. –  Christopher A. Wong Jan 10 '13 at 10:36
Yes,you are right .I have edited my post.Thanks a lot. –  user52976 Jan 10 '13 at 10:48
By the way, the question is phrased really badly. I would phrase it as: "Define the following sequences (for n>=2) by a) ... , b) ... , c) ... , \$d) ... . For which of them does the sum diverge but the sum of the squares converge?" –  Adam Rubinson Jan 10 '13 at 13:15