# Limit question

1. Oct 3, 2007

### azatkgz

lim(n to inf)[e^n-(1+1/n)^n^2]
after some calculations [e^n-e^(n-1/2+1/(3n)+o(1/n))] *
then i took e^n out,so e^n(1-e^(-1/2+1/(3n)+o(1/n))) and i expanded second term
e^n(1-1+1/2-1/(3n)+o(1/n)),so the limit is inf.But my teacher said that my calculations up to * is true then all calculations are wrong.

2. Oct 3, 2007

### morphism

That's hard to read. Why don't you try typing it out in LaTeX?

3. Oct 4, 2007

### SanjeevGupta

looks like you are trying to find
$$\lim_{n \rightarrow\infty} [e^n-(1+\frac{1}{n})^{n^2}}]$$
If so have you tried using the fact that $$n^2=n*n$$
and looking at what happens to
$$\lim_{n \rightarrow\infty} (1+\frac{1}{n})^{n}$$

4. Oct 4, 2007

5. Oct 4, 2007

### azatkgz

$$\lim_{n \rightarrow\infty}[(1+\frac{1}{n})^n}]=e^n$$
than
$$\lim_{n \rightarrow\infty} [e^n-(1+\frac{1}{n})^{n^2}}]=e^n-e^n=0$$

Is it true?

6. Oct 4, 2007

### azatkgz

but my teacher said that my solutions up to
$$e^n-e^{n-\frac{1}{2}+\frac{1}{3n}+o(\frac{1}{n})}$$
Here first part is tends to infinity faster.Then is it infinity?

7. Oct 4, 2007

### SanjeevGupta

I couldn't see how yo got that expansion but even a direct expansion of the bracketed term still gives you just exp(n) so I think the answer is just zero.

8. Oct 5, 2007

### azatkgz

Today my prof.said that the answer is really infinity.

9. Oct 5, 2007

### SanjeevGupta

I would like to see that proof.

10. Oct 5, 2007

### azatkgz

$$e^n-e^{n^2ln(1+\frac{1}{n})}$$

$$e^n-e^{n^2(\frac{1}{n}-\frac{1}{2n^2}+\frac{1}{3n^3}+o(\frac{1}{n^3})}$$

$$e^n-e^{n-\frac{1}{2}+\frac{1}{3n}+o(\frac{1}{n})}$$
this one tends to
$$e^n(1-\frac{1}{\sqrt{e}})$$

Last edited: Oct 5, 2007
11. Oct 5, 2007

### SanjeevGupta

Thank you. Very interesting result and still not obvious to me.

12. Oct 7, 2007

### Gib Z

Quite a clever method.

SanjeevGupta- What he did was simply note that since the exponential and logarithm are inverse functions, $$x = e^{\ln x}$$. Then he also used the result $$\log (k^n) = n \log k$$, and then expanded the log with its taylor series.

13. Oct 7, 2007

### SanjeevGupta

Thanks Gib Z. I now realise that since $$e^n$$ is an infinitely long series to start with and whatever n we chose $$(1+\frac{1}{n})^{n^2}$$ has a finite number of terms so there's always going to be an infinitely long string of terms left after the subtraction starting with the $$n^2+2$$ term of $$e^n$$.
So this reminds me to be very careful when looking at limits of expressions that involve infinite series.