Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
possible duplicate of convergence of a series involving $x^\sqrt{n}$ –  user53153 Jan 15 '13 at 23:29
add comment

3 Answers

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: $$ \sum_{n=1}^\infty a^{\sqrt{n}} \leq \int_0^\infty a^{\sqrt{x}}\,dx = 2\int_0^\infty a^t t\, dt < \infty. $$

share|improve this answer
add comment

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.

share|improve this answer
add comment

iAnother way to look at it is this:

Consider the infinite sum:

$\sum^{n}_{i=0}a^{p}$

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:

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

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.