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I'm trying to prove that $a_n=0.95^{n^2}$ converges to 0 using the formal definition of convergence.

I know that the formal definition of convergence is as follows: given $\varepsilon>0\;\;\exists N$ such that whenever $n>N$ one has $|x_n-L|<\varepsilon$ where $L$ is the limit point.

I'm a bit rusty on these proofs and was wondering if someone could point me to the right direction of finding such an $N$.

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    $\begingroup$ Since that is a decreasing function of $n$ (I figure you are able to prove this), you just solve the inverse problem $a_n = \epsilon$ and take $ N $ to be any integer larger or equal to the solution. $\endgroup$ – derpy Feb 5 at 17:49
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Welp, ....

we want to conclude that $|0.95^{n^2} - 0| < \epsilon$ which (as $0.95^k>0$ always) would be concluded by showing $0 < 0.95^{n^2} < \epsilon$

which will be true if $\ln 0.95^{n^2} = n^2 \ln 0.95< \ln \epsilon$. As $0.95 < 1$ we know $\ln 0.95 < 0$ so this will be true if

$n^2 > \frac {\ln \epsilon}{\ln 0.95}$. The will be a positive value if we choose $\epsilon < 1$.

So if $\epsilon < 1$ then this will be true if $n > \sqrt{\frac {\ln \epsilon}{\ln 0.95}}$.

So if $\epsilon \ge 1$ Then $|0.95^{n^2}-0| < \epsilon$ for all $n > 1$.

If $0< \epsilon < 1$ then for all $n > \sqrt{\frac {\ln \epsilon}{\ln 0.95}}$ we have $|0.95^{n^2} - 0| < \epsilon$

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You could use the following lemma:

If $(a_n)_n$ is such that $0 \leq |a_n| \leq b_n$ for some sequence $(b_n)_n$ that converges to 0, then $(a_n)_n$ also converges to 0.

The latter can be proven using the formal $\varepsilon$ definition.

Finally, choosing $b_n = 0.95^n$ and since $|0.95|<1$, $b_n$ converges to 0.

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  • $\begingroup$ I would just add $0<|a_n|<1$ because we now it is the lower bound. $\endgroup$ – Invisible Feb 5 at 18:26
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Hint Ley $\epsilon >0$. Pick some $A$ such that $0<\frac{1}{A} < \epsilon$.

Now, by Bernoulli inequality $$\frac{1}{(.95)^{n^2}}= \left(1+ (\frac{1}{.95}-1) \right)^{n^2} \geq 1+n^2 (\frac{1}{.95}-1) $$

Make $$1+n^2 (\frac{1}{.95}-1) >A$$

Deduce that that this implies that $$0.95^{n^2}<\epsilon$$

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Another Bernoulli.

Let $r =1/(1+c), c > 0$. Then $(1+c)^m \ge 1+mc > mc$ so $r^m < 1/(mc) < \epsilon$ for $m > 1/(c\epsilon)$.

If $r=.95, c = 1/r-1 = 1/19$. Putting $n^2$ for $m$ gives $n > \sqrt{19/\epsilon}$ as a sufficient bound.

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