$$ \sum_{n=1}^{999} \log_{10}\left(\frac{n+1}{n}\right) $$
Can anybody help me how to calculate this summation?
Anyone knows how to calculate the sum of this series
euler-maclaurinlogarithmssequences-and-seriessummationtaylor expansion
Related Solutions
Hint: This is just a starter which might be helpful for further calculation. We obtain a somewhat simpler representation of the triple sum.
\begin{align*} \sum_{j=1}^m&\sum_{i=j}^m\sum_{k=j}^m\frac{\binom{m}{i}\binom{m-j}{k-j}}{\binom{k}{j}}r^{k-j+i}\\ &=\sum_{j=1}^m\sum_{i=j}^m\sum_{k=j}^{m}\frac{m!}{i!(m-i)!}\cdot\frac{(m-j)!}{(k-j)!(m-k)!}\cdot\frac{j!(k-j)!}{k!}r^{k-j+i}\tag{1}\\ &=\sum_{j=1}^m\sum_{i=j}^m\sum_{k=j}^{m}\frac{\binom{m}{k}\binom{m}{i}}{\binom{m}{j}}r^{k-j+i}\tag{2}\\ &=\sum_{j=1}^m\frac{1}{\binom{m}{j}r^j}\sum_{i=j}^m\binom{m}{i}r^i\sum_{k=j}^m\binom{m}{k}r^k\tag{3}\\ &=\sum_{j=1}^m\frac{1}{\binom{m}{j}r^j}\left(\sum_{k=j}^m\binom{m}{k}r^k\right)^2 \end{align*}
Comment:
In (1) we write the binomial coefficients using factorials
In (2) we cancel out $(k-j)!$ and rearrange the other factorials to binomial coefficients so that each of them depends on one running index only
In (3) we can place the binomial coefficients conveniently and see that the sums with index $i$ and $k$ are the same
METHODOLOGY $1$: Simple Bounds and Using the Squeeze Theorem
Since the arctangent is monotonically increasing, $\arctan(1/\sqrt{x-1})$ is monotonically decreasing. Hence we can assert that
$$\int_{N}^{M+1}\arctan\left(\frac1{\sqrt{x-1}}\right)\,dx\le \sum_{j=N}^M\arctan\left(\frac1{\sqrt{j-1}}\right)\le \int_{N-1}^{M}\arctan\left(\frac1{\sqrt{x-1}}\right)\,dx\tag1$$
The antiderivative of $\arctan\left(\frac1{\sqrt{x-1}}\right)$ can be written as
$$\int \arctan\left(\frac1{\sqrt{x-1}}\right)\,dx=\sqrt{x-1}+x\arctan\left(\frac1{\sqrt{x-1}}\right)+C\tag2$$
For $N=2^n+1$ and $M=2^{n+1}$ in $(2)$, we find that
$$\begin{align} \int_{2^{n}+1}^{2^{n+1}+1} \arctan\left(\frac1{\sqrt{x-1}}\right)\,dx&=2^{n/2}(\sqrt 2-1)\\\\ &+(2^{n+1}+1)\arctan\left(\frac1{\sqrt{2}2^{n/2}}\right)\\\\ &-(2^n+1)\arctan\left(\frac1{2^{n/2}}\right)\tag3 \end{align}$$
Dividing $(3)$ by $2^{n/2}$ and letting $n\to\infty$ reveals that
$$\liminf_{n\to\infty}2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)\ge 2(\sqrt 2-1)\tag4$$
Similarly, evaluation of the right-hand side of $(1)$ reveals
$$\limsup_{n\to\infty}2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)\le 2(\sqrt 2-1)\tag5$$
Finally, putting $(4)$ and $(5)$ together yields the coveted limit
$$\lim_{n\to\infty} 2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)=2(\sqrt 2-1)$$
METHODOLOGY $2$: Use on the Euler-Maclaurin Summation Formula
From the Euler-Maclaurin Summation formula we have
$$\begin{align} 2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)&=2^{-n/2}\int_{2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{x-1}}\right)\,dx+o(1)\\\\ &=2^{-n/2}\left(\sqrt{2^{n+1}-1}-\sqrt{2^n} \right)\\\\ &+2^{n/2+1}\arctan\left(\frac1{\sqrt{2^{n+1}-1}}\right)\\\\ &-(2^{n/2}+1)\arctan\left(\frac1{\sqrt{2^{n}}}\right)+o(1)\tag6 \end{align}$$
Letting $n\to \infty$ in $(6)$, we find that
$$\lim_{n\to\infty} 2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)=2(\sqrt 2-1)$$
Best Answer
You were on the right track! Like Minh Nguyen Nhat, you have to take $\log(1000!)-\log(999!)$ as $\log(1000!/999!)$ to get the answer.
$\log(2/1)+\log(3/2)+\log(4/3)+$...$+\log(1000/999)=\log(1000!/999!)=\log(1000)=3$