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?

  • 2
    $\begingroup$ I really don't understand how you got a sum of logs to evaluate an integral of exponential, but anyway: these problems are usually designed to do the other way around, meaning: one uses the integral to evaluate a series, not what you want...if I understood correctly. $\endgroup$ – DonAntonio Mar 12 '17 at 22:52
  • $\begingroup$ Presumably, OP took $x_k=\log(1+k/n)$ and then he gets $\sum_{k=1}^{n} e^{x_k}(x_{k}-x_{k-1})$, @DonAntonio $\endgroup$ – Thomas Andrews Mar 12 '17 at 22:57
  • $\begingroup$ @ThomasAndrews Thanks, that seems a reasonable assumption. $\endgroup$ – DonAntonio Mar 13 '17 at 8:17

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 $1.$


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).


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

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


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!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.