Calculate a given limit Calculate the following limit:
$$\lim_{n \rightarrow \infty} \frac{3 + 3^{1/2} + 3^{1/3} + \cdots + 3^{1/n} - n}{\ln(n^2 + 1)}$$
I tried to calculate the sum $3 + 3^{1/2} + 3^{1/3} + \cdots + 3^{1/n}$ but got nothing. I currently have no idea what I should do with this limit.
Thank you!
 A: Observe

\begin{align} \frac{3+3^{1/2}+3^{1/3}+\ldots + 3^{1/n} -n}{\ln(n^2+1)}
 = \frac{(e^{\ln 3}-1)+(e^{\frac{1}{2}\ln 3}-1)+\ldots + (e^{\frac{1}{n}\log 3}-1)}{\ln(n^2+1)}. \end{align}

Using the fact that

\begin{align} x\leq e^x-1 \leq x-2x^2 \end{align}

when $0 \leq x\leq 2$, you can show that
\begin{align}
\alpha \ln 3\leq e^{\alpha \ln 3} -1 \leq \alpha \ln 3-2(\alpha \ln 3)^2
\end{align}
for all $0\leq \alpha\leq 1$. Hence it follows

\begin{align} \frac{\ln 3\cdot (1+\frac{1}{2}+\ldots+\frac{1}{n})}{\ln
 (n^2+1)}\leq&\   \frac{(e^{\ln 3}-1)+(e^{\frac{1}{2}\ln 3}-1)+\ldots +
 (e^{\frac{1}{n}\log 3}-1)}{\ln(n^2+1)}\\ \leq&\  \frac{\ln 3\cdot
 (1+\frac{1}{2}+\ldots +\frac{1}{n})-2(\ln 3)^2\cdot(1+\frac{1}{2^2}+\ldots  + \frac{1}{n^2})}{\ln(n^2+1)}. \end{align}

Using the fact that
\begin{align}
\lim_{n\rightarrow \infty}\left(1+\frac{1}{2}+\ldots +\frac{1}{n} - \ln n\right) = \gamma
\end{align}
where $\gamma$ is the Euler-Mascheroni constant i.e. the limit exists, then it follows
\begin{align}
\lim_{n\rightarrow \infty}\frac{\ln 3\cdot (1+\frac{1}{2}+\ldots+\frac{1}{n})}{\ln (n^2+1)}= \lim_{n\rightarrow \infty}\frac{\ln 3\cdot [(1+\frac{1}{2}+\ldots+\frac{1}{n})-\ln n]+\ln 3\cdot \ln n }{\ln (n^2+1)} = \frac{\ln 3}{2}
\end{align}
and
\begin{align}
 &\lim_{n\rightarrow \infty}\frac{\ln 3\cdot (1+\frac{1}{2}+\ldots +\frac{1}{n})-2(\ln 3)^2\cdot(1+\frac{1}{2^2}+\ldots + \frac{1}{n^2})}{\ln(n^2+1)}\\
 &= \lim_{n\rightarrow \infty} \frac{\ln 3\cdot [(1+\frac{1}{2}+\ldots +\frac{1}{n})-\ln n]+\ln 3\cdot \ln n-2(\ln 3)^2\cdot(1+\frac{1}{2^2}+\ldots + \frac{1}{n^2})}{\ln(n^2+1)} = \frac{\ln 3}{2}.
\end{align}
Hence by the squeeze theorem, we have that

\begin{align} \lim_{n\rightarrow \infty}
 \frac{3+3^{1/2}+3^{1/3}+\ldots + 3^{1/n} -n}{\ln(n^2+1)}= \frac{\ln
 3}{2}. \end{align}

A: Hint: Use Taylor expansion to order 1 on $3^{1/k}=\exp(\frac{1}{k} \ln 3)$. You should eventually run into the harmonic series.
A: We can replace $\ln (n^2+1)$ by $\ln n^2 = 2 \ln n.$ Stolz-Cesaro then shows the limit equals
$$\lim_{n\to \infty}\frac{3^{1/(n+1)} - 1}{2\ln (n+1)- 2 \ln n} = \lim_{n\to \infty}\frac{3^{1/(n+1)} - 1}{2\ln (1+1/n)}.$$
Now $\ln (1+1/n) \sim 1/n \sim 1/(n+1),$ so the above equals
$$\lim_{n\to \infty}\frac{1}{2}\frac{3^{1/(n+1)} - 1}{1/(n+1)}.$$
Since $\lim_{h\to 0}(3^h - 1)/h = d\,3^x/dx|_{x=0} = \ln 3,$ the desired limit is $(\ln 3)/2.$
