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How do I calculate the next limit? $$\lim_{n \to \infty}n{(\frac{3^{1+1/2+...+1/(n+1)}}{3^{1+1/2+...+1/n}}-1)}$$ Should I just write it as: $$\lim_{n \to \infty}n{(\frac{3^13^{1/2}...3^{1/(n+1)}}{3^13^{1/2}...3^{1/n}}-1)}$$But in the end i get to $$\lim_{n \to \infty}n{(3^{1/n+1}-1)}$$ That equals to $$\infty0$$ If this is the way, how do I continue it? I know for a fact, that this limit has to be $>1$

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When $h\to 0$, $$ \frac{3^h -1}{h} = \frac{e^{h\ln 3} - e^{0\ln 3}}{h-0} $$ so the limit should make you think of a derivative at $0$.

Now, setting"$h=\frac{1}{n+1}\xrightarrow[n\to\infty]{}0$, you have $$ n\left({3^{\frac{1}{n+1}} -1}\right) = \frac{3^{\frac{1}{n+1}} -1}{\frac{1}{n+1}}\cdot\frac{n}{n+1} $$ and the second factor tends to $1$, so you only care about the limit of the first factor.

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    Soo.. Let's see if i understand this correctly$ n\left({3^{\frac{1}{n+1}} -1}\right) = \frac{3^{\frac{1}{n+1}} -1}{\frac{1}{n+1}}\cdot\frac{n}{n+1}=\frac{3^h-1}{h} $, But how does this help me, it's stil $\frac{0}{0}$ so I cant quite do anything to it.. – 2017-01-08
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    The limit is (by definition) the derivative at $0$ of the function $f$ defined by $f(x) = e^{x\ln 3}$. That derivative is $f'(0) = ?$ – 2017-01-08
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    Ohh, right, thx you so much. So the answer is $\ln{3}$ – 2017-01-08
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    Yes, indeed (advice: you can almost often check on [Wolfram Alpha](http://www.wolframalpha.com/input/?i=limit%20n(3%5E(1%2F(n%2B1))-1)&t=crmtb01) to have empirical confirmation). – 2017-01-08
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Hint

$$3^{\frac{1}{n+1}}-1=e^{\frac{\ln(3)}{n+1}}-1$$

$$\lim_{X\to 0}\frac{e^X-1}{X}=1$$

You will find $\ln(3)$ .

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Consider the general case : $\displaystyle \lim_{n \to \infty}n(a^{1/n} - 1) = \ln(a)$

Let's make a substitution : $\displaystyle b = {a^{1/n}}$, then $\ln(b) = \ln(a)/n$ and out limit is equal $\lim_{b \to 1}\frac{\ln(a)(b-1)}{\ln(b)} = \ln(a)$