Evaluate $\lim_{n \to \infty}\sum_{k= 1}^n (1 + k/n) \left ( \log ( 1 + k/n) - \log (1 + (k-1)/n)\right)$ I got the sum 
$$ \lim_{n \to \infty }\sum_{k= 1}^n (1 + k/n) \left ( \log ( 1 + k/n) - \log (1 + (k-1)/n)\right) $$
while trying to evaluate the integral $\int_0^{\ln 2} e^x dx$ while partitioning the range. How do I evaluate the sum? or alternatively how do I evaluate $\int_0^{\ln 2} e^x dx$ using Lebesgue integral by partitioning the range? 
 A: We have in fact that
$$
\begin{gathered}
  \sum\limits_{1\, \leqslant \,k\, \leqslant \,n} {\left( {1 + k/n} \right)\left( {\ln \left( {1 + k/n} \right) - \ln \left( {1 + \left( {k - 1} \right)/n} \right)} \right)}  =  \hfill \\
   = \ln \prod\limits_{1\, \leqslant \,k\, \leqslant \,n} {\left( {\frac{{n + k}}
{{n + k - 1}}} \right)^{\left( {1 + k/n} \right)} }  =  \hfill \\
   = \ln \left( {\left( {\frac{{\prod\limits_{1\, \leqslant \,k\, \leqslant \,n} {\left( {n + k} \right)^{n + k} } }}
{{\prod\limits_{1\, \leqslant \,k\, \leqslant \,n} {\left( {n + k - 1} \right)} \prod\limits_{1\, \leqslant \,k\, \leqslant \,n} {\left( {n + k - 1} \right)^{n + k - 1} } }}} \right)^{1/n} } \right) =  \hfill \\
   = \ln \left( {\left( {\frac{{\prod\limits_{1\, \leqslant \,k\, \leqslant \,n} {\left( {n + k} \right)^{n + k} } }}
{{\prod\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\left( {n + k} \right)} \prod\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\left( {n + k} \right)^{n + k} } }}} \right)^{1/n} } \right) =  \hfill \\
   = \ln \left( {\left( {\frac{{\left( {2n} \right)^{2n} }}
{{n^n \prod\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\left( {n + k} \right)} }}} \right)^{1/n} } \right) = \ln \left( {\frac{{\left( {2n} \right)^2 }}
{{n\left( {n^{\,\overline {\,n\,} } } \right)^{1/n} }}} \right) = \ln \left( {\frac{{4n}}
{{\left( {\Gamma (2n)/\Gamma (n)} \right)^{1/n} }}} \right) =  \hfill \\
   = \ln \left( {\frac{{4n}}
{{\left( {\frac{{2^{2n - 1} }}
{{\sqrt \pi  }}\Gamma (n + 1/2)} \right)^{1/n} }}} \right) = \ln \left( {\frac{{4n}}
{{4\left( {\frac{1}
{{2\sqrt \pi  }}\Gamma (n + 1/2)} \right)^{1/n} }}} \right) =  \hfill \\
  \mathop  \propto \limits_{n\, \to \,\infty } \ln \left( {\frac{n}
{{\left( {\frac{1}
{{2\sqrt \pi  }}\sqrt {\frac{{2\pi }}
{{n + 1/2}}} \left( {\frac{{n + 1/2}}
{e}} \right)^{n + 1/2} } \right)^{1/n} }}} \right) =  \hfill \\
  \mathop  \propto \limits_{n\, \to \,\infty } \ln \left( {\frac{n}
{{\left( {1/\sqrt 2 } \right)^{1/n} \left( {n + 1/2} \right)}}e} \right)\;\quad \mathop  \propto \limits_{n\, \to \,\infty } 1 \hfill \\ 
\end{gathered} 
$$
where we have used the following identities
Rising Factorial definition: 
$$
\prod\limits_{0\, \leqslant \,k\, \leqslant \,w - 1} {\left( {z + k} \right)}  = z^{\,\overline {\,w\,} }  = \Gamma \left( {z + w} \right)/\Gamma \left( z \right)
$$
Duplication Formula for Gamma
$$
\Gamma \left( {2\,z} \right) = \frac{{2^{\,2\,z - 1} }}
{{\sqrt \pi  }}\Gamma \left( z \right)\Gamma \left( {z + 1/2} \right)
$$
Stirling Asymptotyc formula for Gamma
$$
\Gamma (z) \propto \sqrt {\,\frac{{2\,\pi }}
{z}\,} \left( {\frac{z}
{e}} \right)^{\,z} \quad \left| {\;z\, \to \,\infty ,\;\;\left| {\,\arg (z)\,} \right|} \right. < \pi 
$$
Note the interesting fact that to your sum can be actually given a closed form value (3rd to last row).
A: Lemma: For $0\le u,$
$$u-u^2/2 \le \ln (1+u) \le u.$$
You can prove this by noting it holds for $u=0,$ and then showing that the inequality holds for the derivatives.
Now the given sum equals
$$(1+1/n)\ln(1+1/n) + \sum_{k=2}^{n}(1 + k/n) [\ln (1+k/n)-\ln (1+(k-1)/n].$$
Note $(1+1/n)\ln(1+1/n) \to 0,$ so we can forget about this term. We can write the remaining sum as
$$\sum_{k=2}^{n}(1 + k/n) \ln \left (\frac{1+k/n}{1+(k-1)/n}\right ) = \sum_{k=2}^{n}(1 + k/n) \ln \left (1+\frac{1/n}{1+(k-1)/n}\right ).$$
By the lemma, the last sum is bounded above by
$$\sum_{k=2}^{n}(1 + k/n) \frac {1/n}{1+(k-1)/n}$$ $$ \tag 1= \sum_{k=2}^{n}(1 + (k-1)/n)\frac {1/n}{1+(k-1)/n} + \sum_{k=2}^{n}\frac {1/n^2}{1+(k-1)/n}.$$
The first sum in $(1)$ equals $(n-1)/n \to 1.$ The second sum $\to 0.$
We have shown that our sum is bounded above by an expression that $\to 1.$ Similarly we can use the lemma to bound our sum from below with an expression that also $\to 1.$ By the squeeze theorem, the desired limit is $1.$
A: 
We can rearrange the summand in terms of two telescoping components and a third component that is the kernel of a Riemann sum.  To that end, we now proceed.


Let $a_n=(1+k/n)\left(\log\left(1+\frac kn\right)-\log\left(1+\frac {k-1}n\right)\right)$.  Then, we can write 
$$\begin{align}
a_n&=(1+k/n)\left(\log\left(1+\frac kn\right)-\log\left(1+\frac {k-1}n\right)\right)\\\\
&=\left(\log\left(1+\frac kn\right)-\log\left(1+\frac {k-1}n\right)\right)\\\\
&+\left(\frac kn\log\left(1+\frac kn\right)-\frac{k-1}{n}\log\left(1+\frac {k-1}n\right)\right)\\\\
&-\frac1n\log\left(1+\frac {k-1}n\right) \tag 1
\end{align}$$

Using $(1)$ it is straightforward to see that 
$$\begin{align}
\sum_{k=1}^na_n&=\sum_{k=1}^n(1+k/n)\left(\log\left(1+\frac kn\right)-\log\left(1+\frac {k-1}n\right)\right)\\\\
&=\sum_{k=1}^n\left(\log\left(1+\frac kn\right)-\log\left(1+\frac {k-1}n\right)\right)\\\\
&+\sum_{k=1}^n\left(\frac kn\log\left(1+\frac kn\right)-\frac{k-1}{n}\log\left(1+\frac {k-1}n\right)\right)\\\\
&-\sum_{k=1}^n\frac1n\log\left(1+\frac {k-1}n\right) \\\\
&=2\log(2)-\frac1n\sum_{k=1}^n\log\left(1+\frac {k-1}n\right)\\\\
&=2\log(2)-\frac1n\sum_{k=1}^{n}\log\left(1+\frac {k}n\right)-\frac1n\log(2)\tag 2
\end{align}$$
Note that we could immediately evaluate the limit of $(2)$ in terms of the Riemann sum $\int_0^1 \log(1+x)\,dx=2\log(2)-1$.  However, this tact is circular in nature inasmuch as we are using the sum of interest to evaluate an integral.
Hence, we proceed next by evaluating the sum on the right-hand side of $(2)$ without direct appeal to integration.

Using Stiling's Formula, we can write the sum in $(2)$ as
$$\begin{align}
\sum_{k=1}^n\log\left(1+\frac {k}n\right)&=\log\left(\prod_{k=1}^n\frac{n+k}{n}\right)\\\\
&=\log\left(\frac{(2n)!}{n!\,n^n}\right)\\\\
&\sim \log\left(\frac{4^n\sqrt{n}}{e^n}\right)\\\\
&=n(2\log(2)-1)+\frac12\log(n) \tag 3
\end{align}$$
whence substituting $(3)$ into $(2)$ and letting $n\to \infty$ yields the coveted limit

$$\lim_{n\to \infty}(1+k/n)\left(\log\left(1+\frac kn\right)-\log\left(1+\frac {k-1}n\right)\right)=1$$



NOTE:
If the use of Stirling's Formula is prohibited, then we can make use of the relationship
$$\sum_{k=1}^n k^m=\frac{n^{m+1}}{m+1}+O(n^m) \tag 4$$
which can be shown using induction.

Using $(4)$, it is easy to see that 
$$\begin{align}
\frac1n\sum_{k=1}^{n}\log\left(1+\frac {k}n\right)&=\frac1n\sum_{k=1}^{n}\sum_{m=1}^\infty\frac{(-1)^{m-1}}{m} \left(\frac kn\right)^m\\\\
&=\frac1n\sum_{m=1}^\infty\frac{(-1)^{m-1}}{m} \left(\frac 1n\right)^m\sum_{k=1}^{n}k^m\\\\
&=\frac1n\sum_{m=1}^\infty\frac{(-1)^{m-1}}{m} \left(\frac 1n\right)^m\frac{n^{m+1}}{m+1}+O\left(\frac1n\right)\\\\
&=\sum_{m=1}^\infty\frac{(-1)^{m-1}}{m(m+1)}+ O\left(\frac1n\right)\\\\
&=2\log(2)-1+ O\left(\frac1n\right)
\end{align}$$
and the result follows immediately as expected!
