# Summation splitting: $\sum_{n=1}^{\infty} \frac{n}{3\cdot 5\cdot 7 \dots (2n+1)}$

$$\sum_{n=1}^{\infty} \frac{n}{3\cdot 5\cdot 7 \dots (2n+1)}$$

Can someone please help me in solving this problem, I tried to take the summation of the numerator and denominator individually but my teacher said that it is wrong to do the summation individually, can somebody please explain why is it wrong to take individual summation and please recommend the correct way of solving this problem

• your teacher just reminded you that $1/2+1/3$ is not $2/5$. – G Cab Apr 29 '17 at 9:00
• Your teacher is correct that it is wrong to do the summation individually. Consider a very simple case: $$\sum\limits_{n=1}^2 \frac{x_n}{y_n} \neq \frac{\sum\limits_{i=1}^2 x_i}{\sum\limits_{j=1}^2 y_j}$$ In general: $$\frac{x_1}{y_1}+\frac{x_2}{y_2} \neq \frac{x_1+x_2}{y_1+y_2}$$ There are many examples where the equality does not satisfy, such as the one on the comment above. – projectilemotion Apr 29 '17 at 9:11

## 3 Answers

Hint. One may observe that, for $n\ge1$, \begin{align} \frac {n}{1\cdot 3\cdots (2n+1)} &=\color{red}{\frac12} \cdot \frac {(2n+1)-1}{1\cdot 3\cdots (2n+1)} \\\\&=\color{red}{\frac12} \cdot \frac {1}{1\cdot 3\cdots (2n-1)}-\color{red}{\frac12} \cdot \frac {1}{1\cdot 3\cdots (2n+1)}, \end{align} then, by telescoping terms, one obtains

$$\sum_{n=1}^\infty\frac {n}{1\cdot 3\cdots (2n+1)}=\color{red}{\frac12}.$$

• can you please explain the first step – Gem Apr 29 '17 at 9:20
• multiply and divide by $2 \cdot 4 \cdot \dots \cdot (2n)$ – G Cab Apr 29 '17 at 9:46

Let $x_n$ be the general term. Once you think the series telescopes, you want to write $x_n$ as $f(n) -f(n+1)$ for some function $f$. In particular, when you have a product in the denominator such as in this case, call it $\Pi(n)$ set $$x_n=\frac {An +B}{\Pi (n)}-\frac {A (n+1)+B}{\Pi(n+1)}.$$ In your case you'll get $A=0, B= 1/2$ that is $$x_n=\frac12\left (\frac {1}{3\cdot\cdots(2n-1)}-\frac {1}{3\cdot\cdots(2n+1)}\right).$$If you then look at $S_N=\sum_{n=1}^N x_n$ you'll see that most of the terms cancel out leaving us $S_N=\frac12\left (1- \frac {1}{3\cdot\cdots(2N+1)}\right)$. Let $N\to\infty$ and you're done.

The other answers show how you can get to the result if you don't know it yet, but it's also possible to take the other direction.

Evaluating the first terms, $\dfrac13+\dfrac2{3\cdot5}+\dfrac3{3\cdot5\cdot7}+\dfrac4{3\cdot5\cdot7\cdot9}=\dfrac{472}{945}\approx0.4995$. From this, you might already suspect that the sum is going to equal $\dfrac12$, and from that, look for a way to prove it.

Observe:

\begin{align*} \dfrac13&=\dfrac12-\dfrac16\\&=\dfrac12(1-\dfrac13)\\ \dfrac13+\dfrac2{3\cdot5}&=\dfrac12-\dfrac1{30}\\&=\dfrac12(1-\dfrac1{3\cdot5})\\ \dfrac13+\dfrac2{3\cdot5}+\dfrac3{3\cdot5\cdot7}&=\dfrac12-\dfrac1{210}\\&=\dfrac12(1-\dfrac1{3\cdot5\cdot7})\\ \dfrac13+\dfrac2{3\cdot5}+\dfrac3{3\cdot5\cdot7}+\dfrac4{3\cdot5\cdot7\cdot9}&=\dfrac12-\dfrac1{1890}\\&=\dfrac12(1-\dfrac1{3\cdot5\cdot7\cdot9}) \end{align*}

This is in a form that should allow you to form the hypothesis

$$\sum_{k=1}^{n}\frac{k}{3\cdot5\cdot7\dots(2k+1)}=\dfrac12(1-\dfrac1{3\cdot5\cdot7\dots(2n+1)})$$

which is easily provable by induction and allows the infinite sum to be obtained.