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my textbook says it can be proven in three steps :

1) for very large $ n $, $ a^{2^{n/2}} < 3^{-n} $

2) $ \sum 2^{n}a^{2^{n/2}} < \infty $

3) $ \sum_{n=1} a^{\sqrt{n}} < \sum_{n=0} 2^{n}a^{2^{n/2}} < \infty $

step 1) is trivial since $a^{2^{n/2}}$ decreases exponentially.

step 2) is proven from step 1. the convergence of $ \sum (\frac{2}{3})^n $ is a proof.

I'm stuck at step 3). I think i'm missing something critical here - from my naive understanding, the exponential decrease of $ a^{2^{n/2}} $ should outweigh everything. how can the convergence of $ \sum a^{\sqrt{n}} $ for $ (0<a<1) $ can be proven?

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possible duplicate of convergence of a series involving $x^\sqrt{n}$ – user53153 Jan 15 '13 at 23:29

The idea is: $$ \sum_{n=2^k}^{2^{k+1}-1} a^{\sqrt{n}} \leq \sum_{n=2^k}^{2^{k+1}-1} a^{\sqrt{2^k}} = 2^ka^{\sqrt{2^k}} $$ This is known as Cauchy Condensation Test.

In this situation, the bound given by the Integral Test is somewhat similar: $$ \sum_{n=1}^\infty a^{\sqrt{n}} \leq \int_0^\infty a^{\sqrt{x}}\,dx = 2\int_0^\infty a^t t\, dt < \infty. $$

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We have $\int a^{\sqrt x}\,dx=\dfrac2{(\log a)^2}\,a^{\sqrt x}(\sqrt x\,\log a-1)\xrightarrow{x\to\infty}0$ since $\log a<0$ (verify!). Therefore the series $\sum a^{\sqrt n}$ converges by integral test.

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iAnother way to look at it is this:

Consider the infinite sum:


If p > 1 this converges for all a such that abs(a) < 1 (basic principle behind geometric series)

We note that given your particular equation sqrt(i) > 1 from i = 2 onwards. Which in other words means that:


Converges for abs(a) < 1.

Now if we just evaluate the expression for i = 0 and 1 (which amounts to the values of 1 and a.

Then we can write:

$\sum^{n}_{i=0}a^{i^{1/2}} = 1 + a + \sum^{n}_{i=2}a^{i^{1/2}}$

Of which the all 3 terms are known to converge if abs(a) < 1 thereby implying that:

$\sum^{n}_{i=0}a^{i^{1/2}}$ converges for abs(a) < 1

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