Cfrgtkky

$lim limits_{n to infty } n((n+1)^{frac{1}{100}}-n^frac{1}{100})$

$n((n+1)^{frac{1}{100}}-n^frac{1}{100})=n^frac{101}{100}((1+frac{1}{n})^{frac{1}{100}}-1)=n^frac{101}{100}(e^{frac{ln(1+1/n)}{100}-}-1)$...
hmm, Is this a good approach?

hint

$$(n+1)^a-n^a=n^aBigl((1+frac{1}{n})^a-1Bigr)$$

observe that

$$lim_{nto +infty}frac{e^{aln(1+frac 1n)}-1}{aln(1+frac 1n)}=1$$

and

$$lim_{nto+infty}nln(1+frac 1n)=1$$

You will find $+infty$.

Hint:

Rationalize the numerator

The general term of the denominator will be $$(n+1)^{(99-r)/100}n^{r/100},0le rle99$$

Alternatively,

Set $1/n$ to find
$$lim_{n to infty } n((n+1)^m-n^m=lim_{hto0^+}dfrac{(1+h)^m-1}{h^{1+m}}=lim_{hto0^+}dfrac{mh+binom m2h^{m-1}+cdots+h^m}{h^{1+m}}=+infty$$

Depending on what exactly you're able to use, this certainly could be a good approach. If you know such facts as

$$lnleft(1 + frac 1 nright) ge frac 1 2 cdotfrac 1 n$$

for all sufficiently large $n$ (which is an easy consequence of the fact that $ln(1+x)$ has derivative $1$ at $x = 0$), then you can quickly finish the problem:

$$e^{frac{ln(1+1/n)}{100}} - 1 ge e^{1/(200n)} - 1 ge 1 + frac{1}{200 n} - 1 = frac{1}{200n}.$$
where the inequality is a consequence of the fact that $e^{t} ge 1 + t$. Now combine this with your power of $n$ to see that the limit is infinite.

That being said, I certainly think that binomial expansion is a more natural approach to the problem. But the advantage of your method is that it really only requires two facts: that the derivative of $ln(1+x)$ is $1$ when $x$ is zero, and that the derivative of $exp$ is $1$ when $x$ is zero (plus, of course, the mean value theorem to actually apply this information).

Use binomial expansion

$$(n+1)^{frac{1}{100}}=n^frac{1}{100}(1+1/n)^{frac{1}{100}}=n^frac{1}{100}left(1+frac1{100n}+Oleft(frac1{n^2}right)right)=$$$$=n^frac{1}{100}+frac1{100n^{99/100}}+Oleft(frac1{n^{199/100}}right)$$

and therefore

$$ n((n+1)^{frac{1}{100}}-n^frac{1}{100})=frac{n^{frac1{100}}}{100}+Oleft(frac1{n^{99/100}}right)toinfty$$