What is the Convergence of the Infinite Sum with k^(1/k)?

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Homework Help Overview

The discussion revolves around the convergence of the infinite sum \(\sum_{k=0}^{\infty} \sqrt[k]{k} - 1\). Participants are exploring methods to demonstrate convergence, particularly focusing on the behavior of the general term as \(k\) approaches infinity.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the limit of the general term \(a_k = \sqrt[k]{k} - 1\) and its behavior as \(k\) increases. There is mention of using the ratio test, comparison with other sums, and asymptotic comparison methods. Some participants express uncertainty about where to begin and seek simpler methods.

Discussion Status

There is an ongoing exploration of different approaches to assess convergence. A hint has been provided regarding the limit of the general term and its implications for establishing convergence. Some participants are considering the asymptotic comparison test as a viable direction.

Contextual Notes

One participant indicates a preference for simpler methods, suggesting that the complexity of the proposed approaches may be a barrier to understanding. The original poster expresses a lack of confidence in their initial attempts.

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Homework Statement



Show that
\sum_{k=0}^{\infty} \sqrt[k]k-1

converges.

Homework Equations



Ratio, radix theorems, comparison with other sums...


The Attempt at a Solution



No idea whatsoever.
Where does one begin in this case ? With other cases I'm quite confident.
 
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Hint:
Let the general term be denoted by:
<br /> a_k = \sqrt[k]{k} - 1<br />
Do you know how to prove that \lim_{k \rightarrow \infty}{a_k} = 0?
Then, we have:
<br /> k = (1 + a_k)^{k}<br />
Taking the same equality for k + 1, and subtracting this one, you ought to get:
<br /> 1 = (1 + a_{k + 1})^{k + 1} - (1 + a_k)^{k}<br />
Solve this equation for a_{k + 1}. What do you get?

Because as k \rightarrow \infty, a_k is an infinitesimal quantity, you may expand your expression for a_{k + 1} in powers of a_{k} up to the first non-vanishing order. What do you get?

The solution for this asymptotic recursion relation would give you a comparison general term b_n, and the series \sum_{n = 1}^{\infty}{b_n} is pretty easy to test for convergence. Then, you may use the [STRIKE]Ratio comparison test[/STRIKE].

EDIT:
Use the Asymptotic comparison test mentioned here instead.
 
Last edited:
I don't quite follow you.

I sort of need simpler methods, thanks anyway.
 
Yep, asymptotic comparison seems a good way.
 

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