# Is $\{x_n\}$ convergent?

Let $$x_n=\sum_{k=1}^{n^2}\frac{1}{\sqrt{n^4+k}}$$ Is $\{x_n\}$ convergent? If it is convergent then find its limit.

Here n- th term of the series $\frac{1}{\sqrt{n^4+n^2}} \to 0$ as $n \to \infty$. Can I conclude from here that $\{x_n\}$ is convergent?

Again $$\lim_{n\to \infty }x_n=\lim_{n\to \infty }\sum_{k=1}^{n^2}\frac{1}{\sqrt{n^4+k}}=\lim_{m\to \infty }\frac{1}{m} \sum_{k=1}^{m}\frac{1}{\sqrt{1+\frac{k}{m^2}}}$$Then I am stuck .Please help.

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Keep in mind that just because $a_n\to 0$, that does not mean $\sum a_n$ converges. (Just consider $a_n=1/n$.) –  JohnD Jan 26 '13 at 15:48

## 3 Answers

For each term of the sum, we have $$\frac{1}{\sqrt{n^4+n^2}} \leq \frac{1}{\sqrt{n^4+k}}\leq \frac{1}{\sqrt{n^4+1}}.$$ Summing these inequalities from $k=1$ to $k=n^2$ yields $$\frac{1}{\sqrt{1+1/n^2}}=\frac{n^2}{\sqrt{n^4+n^2}} \leq x_n \leq \frac{n^2}{\sqrt{n^4+1}}=\frac{1}{\sqrt{1+1/n^4}} .$$ Now both ends converge to $1$.

Therefore, by the Squeeze Theorem, $\lim_{n\rightarrow +\infty} x_n=1$.

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Having $\lim_{x\rightarrow\infty}x_n = 0$ is a necessary condition but not a sufficient one - See the Harmonic series for ex, $\sum \frac{1}{n}$ which doesn't converge.

I know it's a clumsy way to do it, but would the following be good?

$$\lim_{L\rightarrow\infty} \int_1^L \frac{1}{\sqrt{L^2+x}} = \lim_{L\rightarrow\infty} 2\left( \sqrt{L^2+L} - \sqrt{L^2+1} \right) = \lim_{L\rightarrow\infty} 2\frac{L-1}{\sqrt{L^2+L} + \sqrt{L^2+1}} = 1$$

and conclude that by the integral test it converges?

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We can use your simplification, followed by Squeezing.

Note that $$1\lt \sqrt{1+\frac{k}{m^2}}\lt 1+\frac{k}{2m^2}.$$ Thus if $k\le m$, then $$\frac{1}{\sqrt{1+\frac{k}{m^2}}} \ge \frac{1}{1+\frac{m}{2m^2}}.$$

It is simpler not to use the simplification, and note directly that $\sqrt{n^4+k}\le \sqrt{n^4+n^2}\lt n^2+\frac{1}{2}$.

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