# Yet another limit to a sequence

so, I'm completely stuck with this limit:

$$\lim_{n\to\infty}{\left(1+\frac{1}{n^2}\right)^n}$$

Which I can't even grasp how to start with, since I didn't understand the explanation of some method having to do with $e$. However, I believe that it would go something like this:

$$\lim_{n\to\infty}{\left(1+\frac{1}{n^2}\right)^{n^2}}$$

$$\lim_{n\to\infty}{\left(\left(1+\frac{1}{n^2}\right)^{n^2}\right)^{1/n^2I}}$$

... Or something like that... However, as you can see, I'm completely blank.

Any suggestions will be appreciated.

-
You're basically there; just get your algebra right. It's equal to $\lim_{n\rightarrow\infty} \left((1 + 1/n^2)^{n^2}\right)^{(1/n)}$. The inner part goes to $e$, and the remaining exponent ($1/n$) goes to $0$, so the whole thing goes to $e^{0}=1$. – mjqxxxx Sep 11 '12 at 5:39

Yes, something like that. The following rewrite will work: $$\left(\left(1+\frac{1}{n^2}\right)^{n^2}\right)^{1/n}.$$

We know that as $n\to\infty$, $(1+1/n^2)^{n^2}$ approaches $e$. In particular, if $n$ is large enough (and it doesn't need to be large), $(1+1/n^2)^{n^2}$ is between $1$ and $4$. Now take the $n$-th root. The result is between $1$ and $4^{1/n}$. It is a standard result that if $a$ is positive, then $a^{1/n}$ approaches $1$ as $n\to\infty$. So by squeezing our limit is $1$.

Another way: Alternately, you could look at $(1+1/x^2)^x$ as $x$ gets large. Calculate the logarithm, which is $x\log(1+1/x^2)$, and use L'Hospital's Rule. If I were using this approach, I would probably make the substitution $y=1/x$, and find the limit as $y$ approaches $0$ from the right. So we want to find $$\lim_{y\to 0^+}\frac{\ln(1+y^2)}{y}.$$ Easily the limit turns out to be $0$. Now exponentiate.

If you are familiar with power series, you can find the Taylor expansion of $\ln(1+y^2)$ instead.

-
+1 for not needing the e limit. – Hagen von Eitzen Sep 11 '12 at 6:31

You know that $\lim\limits_{n\to\infty}\left(1+\frac1n\right)^n=e$. Since $n^2\to\infty$ as $n\to\infty$, there is no harm in replacing $n$ by $n^2$: $$\lim_{n^2\to\infty}\left(1+\frac1{n^2}\right)^{n^2}=e\;.\tag{1}$$ And again because $n^2\to\infty$ when $n\to\infty$ and vice versa, $(1)$ says the same thing as $$\lim_{n\to\infty}\left(1+\frac1{n^2}\right)^{n^2}=e\;.\tag{2}$$

Now your limit looks a lot like $(2)$, except that it has $n$ in the exponent instead of $n^2$, so we apply a little algebraic judo to $(2)$ to rewrite it as

$$\left(1+\frac1{n^2}\right)^n=\left(\left(1+\frac1{n^2}\right)^{n^2}\right)^{1/n}$$

and take limits as $n\to\infty$:

$$\lim_{n\to\infty}\left(1+\frac1{n^2}\right)^n=\lim_{n\to\infty}\left(\left(1+\frac1{n^2}\right)^{n^2}\right)^{1/n}\;.$$

The bit inside the inner parenthesis now behaves in a known way $-$ see $(2)$ $-$ as $n$ gets larger and larger, and you can use that behavior to finish the job.

-