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I was trying to prove the convergence of the series $\sum_{n=1}^{\infty}x^{\sqrt{n}}$, for $0<x<1$. Unfortunately, I could not make one of the standard convergence tests give me an answer. Does anybody of you have a suggestion? any help is much appreciated!

many thanks!

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Are you sure the integral test doesn't work? I think it does... Write x^(sqrt(n)) = exp[(n^1/2)ln(x)]. Then for any 0<x<1 x is a constant and so ln(x) < 0 is a constant. Let u = n^1/2 and do some substitutions and Bob's your uncle – Adam Rubinson Jan 5 '13 at 20:53
What about the root test? – user39097 Jan 5 '13 at 21:02
Neither the ratio test, nor the root test will help here. – 1015 Jan 5 '13 at 21:06
@user39097 Root Test fails because when you take the $n$th root, you get $x^{1/\sqrt{n}}$ which goes to $1$ as $n\to\infty$, which makes the Root Test inconclusive. – Joe Jan 5 '13 at 21:09
up vote 2 down vote accepted

Let $x=\dfrac{1}{e^y}$. Note that $y$ is positive. There is an integer $k$ such that $ky\gt 2$.

For large enough $n$, $\sqrt{n}\gt k\log n$. We can if we wish prove this using L'Hospital's Rule, by showing that $\lim_{n\to\infty} \frac{\log n}{\sqrt{n}}=0$;

For any such $n$. we have $$x^{\sqrt{n}}=\frac{1}{e^{y\sqrt{n}}}\lt \frac{1}{e^{ky\log n}}=\frac{1}{n^{ky}}\lt \frac{1}{n^2}.$$

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thanks guys for all your great answers, I really appreciate your support!!! – s_2 Jan 5 '13 at 21:56

Let us compute $$ n^2x^\sqrt{n}=e^{2\ln n+\ln x\sqrt{n}}=e^{\ln x \sqrt{n}\left(1+\frac{2\ln n}{\ln x\sqrt{n}}\right)}. $$ Thus we see that $$ \lim_{n\rightarrow +\infty} n^2x^\sqrt{n}=0. $$ So the sequence $(n^2x^\sqrt{n})$ is bounded and there exists $C>0$ such that $$ 0\leq x^\sqrt{n}\leq \frac{C}{n^2} $$ for all $n\geq 1$. Then the convergence of $\sum_{n\geq 1} x^\sqrt{n}$ follows by comparison.

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+1, but you might want to add some insight on why you chose $n^2$ to help you find your upper bound – Joe Jan 5 '13 at 21:05
Before I do the ratio test or the root test, I always compare a series with positive general term with the famous family of convergent series $\sum_{n\geq 1} 1/n^\alpha$ (for $\alpha >1$). – 1015 Jan 5 '13 at 21:10
In fact you could use $\sum 1/n^a$ for any $a$ ... your general term is smaller than all of these. – GEdgar Jan 5 '13 at 21:24
@GEdgar Yes, indeed. – 1015 Jan 5 '13 at 21:29

For $x \in (0,1)$, we have $$x^{\sqrt{n}} \leq x^k \,\,\,\,\,\,\,\,\,\,\ \forall n \geq k^2$$ $$\sum_{n=k^2}^{(k+1)^2-1}x^{\sqrt{n}} \leq 2kx^k$$ Hence, we have $$\sum_{n=1}^{(N+1)^2-1} x^{\sqrt{n}} \leq \sum_{n=1}^N 2nx^n$$ Hence, $$\sum_{n=1}^{\infty} x^{\sqrt{n}} $$ converges for $x \in (0,1)$ since $ \displaystyle \sum_{n=1}^\infty 2nx^n$ converges for $x \in (0,1)$.

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As Adam mentioned, the integral test will work. Just have to be careful that everything being done with respect to $n$ and remember that $0<x<1$. Let's fix $x$ between zero and one and we see that $f(n)=x^{\sqrt{n}}$ is non-negative and monotonically decreasing so we can use the integral test.

We need $$\int_1^{\infty} x^{\sqrt{n}} dn = \int_1^{\infty} e^{\sqrt{n} \ln(x)}dn.$$

Remember that $\ln(x)$ is a negative constant here so the integral does converge. But if you want to go ahead and explicitly show it, you can do it. Just use the substitution $$u = \sqrt{n}$$ $$ du = \frac{dn}{2\sqrt{n}}$$ $$2udu=dn$$

and then integrate by parts and undo the substitution and you get

$$\left.\frac{2x^{\sqrt{n}}(\sqrt{n} \ln(x)-1)}{(\ln(x))^2}\right]_{n=1}^{\infty}$$

which goes to zero as $n\rightarrow\infty$ (because $0<x<1$) and plugging in $n=1$ simplifies it so you get $$0-\frac{2x(\ln(x)-1)}{(\ln(x))^2}$$ which of course is finite for our fixed $x$. Therefore the original series converges because this integral converges.

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Try the integral test ! $\quad\displaystyle\sum_{n=1}^\infty a^{-\sqrt[k]n}~=~-1+\sum_{n=0}^\infty a^{-\sqrt[k]n}\quad{\large\approx}~-1+\int_0^\infty a^{-\sqrt[k]t}~dt~=~\frac{k!}{\ln^ka}-1.$

See $\Gamma$ function for more details. In this case, $a=\dfrac1x$ and $k=2$.

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