# Convergence of $\sum\limits_{n=0}^{\infty}\frac{n-\sqrt{n}}{(n+\sqrt{n})^2}$

I want to check, whether $\sum\limits_{n=0}^{\infty}\frac{n-\sqrt{n}}{(n+\sqrt{n})^2}$ converges or diverge.

I tried to use the Ratio test:

\begin{align} \left|\dfrac{a_{n+1}}{a_n}\right| & = \dfrac{n+1-\sqrt{n+1}}{(n+1+\sqrt{n+1})^2}\dfrac{(n+\sqrt{n})^2}{n-\sqrt{n}}\\ & = \dfrac{-1}{n+1+\sqrt{n+1})} (-n-\sqrt{n})\\ & = \dfrac{n+\sqrt{n}}{n+1+\sqrt{n+1})} \leq 1 \end{align}

So $\displaystyle \sum\limits_{n=0}^{\infty}\frac{n-\sqrt{n}}{(n+\sqrt{n})^2}$ converges.

Could somebody please check my solution?

• For $a_n = \frac1n$, you also have $\left\lvert \frac{a_{n+1}}{a_n}\right\rvert < 1$. That the ratio is smaller than $1$ is not sufficient. (And indeed, the comparison test shows that your series diverges.) – Daniel Fischer Dec 6 '13 at 19:46
• Could anybody tell me, where my mistakes are? – fear.xD Dec 6 '13 at 19:53
• You were supposed to test $\lim \frac{a_{n+1}}{a_n}$. – Karolis Juodelė Dec 6 '13 at 19:54
• You need to find the limit not just the ratio. – hhsaffar Dec 6 '13 at 19:57
• @fear.xD : there are at least two mistakes: (1) concluding that if the limit is $\leq 1$, then the series converges; and (2) using the Ratio Test in the first place. Well, (2) is not exactly a mistake, just a waste of time. With a little more experience you will recognize a series like this as a bad candidate for the Ratio Test. – Stefan Smith Dec 7 '13 at 0:00

Consider strictly $n\geq 4$ (this won't affect the convergence of the series). Then $\sqrt{n}\leq \frac n2$ (why?); this means that you can say that $n-\sqrt{n}\geq \frac n2$. Likewise, you have that $n+\sqrt{n}\leq n+\frac n2 = \frac{3n}2$. Combine these and you get that $\frac{n-\sqrt{n}}{\left(n+\sqrt{n}\right)^2}\geq \dfrac{\frac n2}{\left(\frac{3n}2\right)^2} = \frac{2}{9n}$. Now use the divergence of the harmonic series.

The problem with your proposed solution is that you've misapplied the ratio test; you need to be able to say that $\left(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\right)\lt 1$, not just that the limit is $\leq 1$ or even that $\left|\frac{a_{n+1}}{a_n}\right|\lt 1$ for all $n$. In fact, the harmonic series $\sum_{n\to\infty}\frac1n$ itself serves as a counterexample to your use of the limit test; with $a_n=\frac1n$, we have $\left|\frac{a_{n+1}}{a_n}\right|\lt 1$ for all $n$, $\lim_{n\to\infty} a_n=0$, and $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\leq 1$, but $\sum a_n$ diverges. In general, the harmonic series can serve as a useful general-purpose counterexample: whenever you're unsure of your application of a series convergence test, try asking yourself 'what happens if I apply my logic to the harmonic series?'.

You can use the limit comparison test with the harmonic series $\sum_{n=1}^\infty \frac1n$. Since $$\lim_{n\to\infty} \frac{(n-\sqrt n)/(n+\sqrt n)^2}{1/n} = 1$$ is finite and nonzero, either both series converge or both diverge. The harmonic series definitely diverges, and so $\sum_{n=1}^\infty \frac{n-\sqrt n}{(n+\sqrt n)^2}$ also diverges.

(As Daniel mentioned in a comment, the regular comparison test with the harmonic series also works here. But the limit comparison test is very valuable, as we don't need to worry about actual inequalities, only limiting values.)

• Agreed, although with a caveat: anyone with a working knowledge of orders of growth 'knows' that the limit of the ratio is 1 almost trivially, but showing that it's actually equal to 1 is another several lines of error-prone algebra for most students. – Steven Stadnicki Dec 6 '13 at 21:27
• I suspect it's more straightforward/trainable algebra though; once you successfully write $$\frac{(n-\sqrt n)/(n+\sqrt n)^2}{1/n} = \frac{1-1/\sqrt n}{(1+1/\sqrt n)^2},$$ you're home free. – Greg Martin Dec 6 '13 at 23:58

Considering dominant terms, the numerator behaves like $n$ and the denominator behaves like $n^2$. $n/n^2 = 1/n$ and the harmonic series diverges, so the series will diverge. In problems like this involving only powers of $n$, the Ratio Test and Root Test are nearly always inconclusive. The Limit Comparison Test (comparing to $1/n$) will work.

By the way, for the Ratio Test, the limit you were doing needs to be strictly less than $1$, not just $\leq 1$. So your conclusion is unsupported by your work. I didn't look too closely at your work because I know the Ratio Test will not work on this problem.