I am finding the positive values of $x$ for which the following series is convergent $$ \sum_{n=1}^{\infty}x^{\sqrt{n}}$$ It is sure that it is not convergent for $x\geq1$ as $n$-th term will not tend to zero. Now $x\in[0,1)$ how to check its convergence? Please help me to solve it. Thanks.
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4For $x\geqslant1$, $x^{\sqrt{n}}$ does not converge to $0$ hence the series diverges. For $x$ in $(0,1)$, using the bound $x^{\sqrt{n}}\leqslant x^k$ for every $k^2\leqslant n<(k+1)^2$ and every $k\geqslant1$ yields $$\sum_{n=1}^\infty x^{\sqrt{n}}\leqslant\sum_{k=1}^\infty(2k+1)x^k,$$ which should allow you to conclude. – Did Sep 30 '15 at 11:56
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right hand series $\sum (2N+1)x^{N}$ is convergent? – neelkanth Sep 30 '15 at 12:02
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and $n$ is not fixed ....how its converges? – neelkanth Sep 30 '15 at 12:05
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"right hand series ∑(2N+1)xN is convergent?" Yes. "and n is not fixed ....how its converges?" What? – Did Sep 30 '15 at 12:13
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1See also: Does the series $\sum_{n=1}^{\infty}|x|^\sqrt n$ converge pointwise? If it then what would be the sum? – Martin Sleziak Aug 05 '22 at 10:22
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@MartinSleziak ok thank you ..... – neelkanth Aug 05 '22 at 11:10
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Without the exponential:
If $0\le x <1$, we have: \begin{align*} \sum_{n\ge1}x^{\sqrt n}&\le\sum_{n\ge1}x^{\lfloor\sqrt n\rfloor}=3x+5x^2+7x^3+9x^4+\dotsm \\ &\le 2+4x+6x^2+8x^3+10x^4+\dotsm=2(1+2x+3x^2+4x^3+5x^4+\dotsm)\\ &= 2\biggl(\frac1{1-x}\biggr)'=\frac2{(1-x)^2} \end{align*} hence the series converges.
Bernard
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Let $x \in ]0,1[$; then $$x^{\sqrt{n}} = \exp(\sqrt{n} \log x) = \frac{1}{e^{\sqrt{n}|\log x|}} < \frac{1}{(\sqrt{n})^{3}} $$ for large $n$, so by the comparison test the desired series converges.
Yes
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For all $a,b > 0$ we have $\xi^{a}/e^{b\xi} \to 0$ as $\xi \to \infty$, @neela – Yes Sep 30 '15 at 12:10
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@neela: Thanks; it makes exponential appear in the denominator, instead of the numerator. – Yes Sep 30 '15 at 12:17
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