# Show that $a_n=0.95^{n^2}$ converges to 0 using the formal definition of convergence

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 $$N$$.

• 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. – derpy Feb 5 at 17:49

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 $$|0.95^{n^2} - 0| < \epsilon$$

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.

• I would just add $0<|a_n|<1$ because we now it is the lower bound. – Invisible Feb 5 at 18:26

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 $$0.95^{n^2}<\epsilon$$

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.