Calculating the Limit of [e^n-(1+1/n)^n^2]

[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.

 

[morphism]

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

 

[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}

 

[azatkgz]

than actually answer is 0.Yes?

 

[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?

 

[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?

 

[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.

 

[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?

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

 

[SanjeevGupta]

I would like to see that proof.

 

[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}})

 

[SanjeevGupta]

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

 

[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.

 

[SanjeevGupta]

Thanks Gib Z. I now realize 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.