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How can I proof the convergence of

$${\{7^{-n^{-1/5}}\}_{n\geq 1}}$$

?

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Take logarithm to base 7, of the sequence $-n^{-1/5}\log 7$ and then it's limit. It'll equal the logarithm to the base 7, of the limit of the sequence because of continuity of $\log_7x$. Let the limit be $L$. Of course you can explicitly calculate $L$. $7^L$ is the required limit. –  Aneesh Karthik C Jul 10 '12 at 12:18
    
The limit of the sequence is 1, but I have to prove the limit. –  hinafu Jul 10 '12 at 12:22
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2 Answers

up vote 3 down vote accepted

Just do some computations and use the motonicity of powers and the logarithm to find the $N$ you want:

First note, that $7^{-n^{-1/5}}\le 1$ for every $n$, now let $\epsilon > 0$. We have \begin{align*} 1 - 7^{-n^{-1/5}} < \epsilon &\iff 7^{-n^{-1/5}} > 1 - \epsilon\\ &\iff -n^{-1/5} > \log_7(1-\epsilon)\\ &\iff n^{-1/5} < \log_7\frac 1{1 - \epsilon}\\ &\iff n^{1/5} > \frac 1{\log_7 (1-\epsilon)^{-1}}\\ &\iff n > \left(\frac 1{\log_7 (1-\epsilon)^{-1}}\right)^5 \end{align*} If we let $N := 1 + \left\lceil (\frac 1{\log_7 (1-\epsilon)^{-1}})^5\right\rceil$, then $\left|{7^{-n^{-1/5}} -1}\right| < \epsilon$ for $n \ge N$.

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Thanks!, I will have to practice a lot from now –  hinafu Jul 10 '12 at 13:00
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If you're trying to prove the convergence of the given sequence, epsilon-$N$ proofs are not the only way. Sometimes, the simplifications and finding the right $N$ may be difficult. Of course if you don't know the limit of the sequence it is of no use.

I will introduce a different approach to this problem.

Consider the two functions $f:[0, +\infty] \to \mathbb{R}$ and $g:[0, +\infty] \to \mathbb{R}$ given by $f(x)=( \frac{1}{7} )^x$ and $g(x)=\sqrt[5]{x}$. Then $g(f(x))=(\frac{1}{7} )^\sqrt[5]{x}=7^{(-\sqrt[5]{x})}$ is well defined for all $x \in [0, +\infty] $ and since $f, g$ are right continuous at $0$, the limit $\lim\limits_{x \to 0+}g(f(x))$ exists and equal to $g(f(0))=7^0=1$.

It is easy to see that this result can be used to prove what you wanted to prove. We know $x \to 0+$ if and only if $ \frac{1}{x} \to +\infty$. Therefore from above we have, $\lim\limits_{x \to +\infty}7^{(-\sqrt[5]{1/x})}=\lim\limits_{x \to +\infty}7^{(-x^{(-1/5)})}=1.$

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