# An Elementary Summation Problem with $\varphi(4,n)$

The complete proof is given by Ramanujan.\begin{align*}\varphi(4,n) & =1+\sum\limits_{k=1}^n\left\{\frac 1{4k-1}+\frac 1{4k+1}-\frac 1{2k}\right\}\\ & =\sum\limits_{k=1}^{4n+1}\frac 1k-\frac 12\sum\limits_{k=1}^{2n}\frac 1k-\frac 12\sum\limits_{k=1}^n\frac 1k\\ & =\sum\limits_{k=1}^{3n+1}\frac 1{n+k}-\sum\limits_{k=1}^n\frac 1{2n+2k}\tag{1.1}\\ & =\sum\limits_{k=1}^n\frac 1{n+k}+\sum\limits_{k=0}^n\frac 1{2n+2k+1}\tag{1.2}\end{align*}

And using the second equality in the above equations, we also find that:\begin{align*}\varphi(4,n) & =\sum\limits_{k=1}^{4n+1}\frac 1k-2\sum\limits_{k=1}^{2n}\frac 1{2k}+\frac 12\sum\limits_{k=1}^{2n}\frac 1k-\sum\limits_{k=1}^n\frac 1{2k}\tag{2.1}\\ & =\sum\limits_{k=1}^{4n+1}\frac {(-1)^{k+1}}k+\frac 12\sum\limits_{k=1}^{2n}\frac {(-1)^{k+1}}k\tag{2.2}\end{align*}

Questions:

1. How did Ramanujan get from $(1.1)$ to $(1.2)$?
2. How do you get the second set of equations?
3. What was Ramanujan's thinking when he split$$-\frac 12\sum\limits_{k=1}^{2n}\frac 1{k}=-2\sum\limits_{k=1}^{2n}\frac 1{2k}+\frac 12\sum\limits_{k=1}^{2n}\frac 1k$$

On $(1.1)$

I've spent some time on this, and I am just confused. I understood every step up until $(1.1)$ and $(2.1)$. If we expand $(1.2)$, then we see that$$\sum\limits_{k=1}^n\frac 1{n+k}+\sum\limits_{k=0}^n\frac 1{2n+2k+1}=\left(\frac 1{n+1}+\frac 1{n+2}+\cdots+\frac 1{2n}\right)+\left(\frac 1{2n+1}+\frac 1{2n+3}+\cdots+\frac 1{4n+1}\right)$$And comparing to the line before that,$$\sum\limits_{k=1}^{3n+1}\frac 1{n+k}-\sum\limits_{k=1}^n\frac 1{2n+2k}=\left(\frac 1{n+1}+\frac 1{n+2}+\cdots+\frac 1{4n+1}\right)-\left(\frac 1{2n+2}+\frac 1{2n+4}+\cdots+\frac 1{4n}\right)$$Which is obviously missing a couple of terms. I'm thinking that Ramanujan added a summation, and subtracted, then combined the summations to get $(1.2)$. But I'm not sure how.

On $(2.1)$

I just don't know where to start. I see that Ramanujan broke up the summation from $k=1$ to $2n$, but I don't see how from that, you can derive the second equation.

I'm self taught. So sorry if these problems seem too elementary and basic. I'm still in the learning phase!

How did Ramanujan get from $(1.1)$ to $(1.2)$?

Well, I can't promise this is how Ramanujan personally came up with it, but here's a derivation:

\begin{align} \color{red}{\sum_{k=1}^{3n+1}\left(\frac 1{n+k}\right)}-\color{blue}{\sum_{k=1}^n\left(\frac 1{2n+2k}\right)} &= \color{red}{\left[\sum_{k=1}^{n}\left(\frac 1{n+k}\right) + \sum_{k=n+1}^{3n+1}\left(\frac 1{n+k}\right)\right]}-\color{blue}{\sum_{k=1}^n\left(\frac 1{2n+2k}\right)}\,\textrm{split red sum into pieces}\\ &=\sum_{k = 1}^n\left( \color{red}{\frac 1{n+k}} - \color{blue}{\frac 1{2n+2k}}\right)+ \color{red}{\sum_{k = n+1}^{3n + 1}\left(\frac 1{n+k}\right)}\\ &\textrm{pair first red sum and blue sum because the values of }k\textrm{ that they're summing over are the same}\\ &= \color{green}{\sum_{k = 1}^n\left(\frac 1{2n+2k}\right)}+ \color{red}{\sum_{k = n+1}^{3n + 1}\left(\frac 1{n+k}\right)}\,\textrm{red and blue combine to green}\\ &= \color{green}{\sum_{k = 1}^n\left(\frac 1{2n+2k}\right)}+\color{red}{\sum_{j = 0}^{2n}\left(\frac 1{n+j + n + 1}\right)}\\ &\textrm{reindex red sum as sum over }j\textrm{ by substituting }k = j + n + 1\\ &= \color{green}{\sum_{k = 1}^n\left(\frac 1{2n+2k}\right)}+\color{red}{\sum_{k = 0}^{2n}\left(\frac 1{2n+k+1}\right)}\\ &\textrm{the name of the dummy index doesn't matter, so change }j\textrm{ to }k\textrm{ in red sum}\\ &= \color{green}{\sum_{k = 1}^n\left(\frac 1{2n+2k}\right)}+ \left[\color{red}{\sum_{\substack{0\leq k\leq 2n\\k\textrm{ even}}}\left(\frac 1{2n+k + 1}\right)} + \color{blue}{\sum_{\substack{0\leq k\leq 2n\\k\textrm{ odd}}}\left(\frac 1{2n+k + 1}\right)}\right]\\ &\textrm{split up red sum into the sum over even }k\textrm{ (red) plus the sum over odd }k\textrm{ (blue)}\\ &= \color{green}{\sum_{k = 1}^n\left(\frac 1{2n+2k}\right)}+ \left[\color{red}{\sum_{k = 0}^n\left(\frac 1{2n+(2k)+ 1}\right)} + \color{blue}{\sum_{k = 1}^{n}\left(\frac 1{2n+ (2k - 1) + 1}\right)}\right]\\ &\textrm{even numbers are those of the form }2i\textrm{ and odd numbers are those of the form }2i - 1.\\ &\textrm{Use these to index even and odd sums explicitly}\\ &= \color{green}{\sum_{k = 1}^n\left(\frac 1{2n+2k}\right)}+ \left[\color{red}{\sum_{k = 0}^n\left(\frac 1{2n+2k+ 1}\right)} + \color{blue}{\sum_{k = 1}^{n}\left(\frac 1{2n+ 2k }\right)}\right]\,\textrm{simplify}\\ &= (\color{green}{1} + \color{blue}{1})\sum_{k = 1}^n\left(\frac 1{2n+2k}\right) + \color{red}{\sum_{k = 0}^n\left(\frac 1{2n+2k+ 1}\right)}\\ &\textrm{green and blue sums are the same, so combine them}\\ &= \sum_{k = 1}^n\left(\frac 2{2n+2k}\right)+ \color{red}{\sum_{k = 0}^n\left(\frac 1{2n+2k+ 1}\right)}\\ &2(a_1 + a_2 + \dots + a_n) = (2a_1 + 2a_2 + \dots + 2a_n)\textrm{ (i.e., move the }2\textrm{ to the inside of the sum)}\\ &= \sum_{k=1}^n\left(\frac 1{n+k}\right)+\color{red}{\sum_{k=0}^n\left(\frac 1{2n+2k+1}\right)}\,\textrm{simplify}. \end{align}

How do you get the second set of equations?

\begin{align*} \sum\limits_{k=1}^{4n+1}\frac 1k-\frac 12\sum\limits_{k=1}^{2n}\frac 1k-\frac 12\sum\limits_{k=1}^n\frac 1k &= \sum\limits_{k=1}^{4n+1}\frac 1k-\sum\limits_{k=1}^{2n}\frac{1}{2k}-\sum\limits_{k=1}^n\frac{1}{2k} \\ &= \sum\limits_{k=1}^{4n+1}\frac{1}{k} + (1 - 2)\sum\limits_{k=1}^{2n}\frac{1}{2k}-\sum\limits_{k=1}^n\frac{1}{2k}\\ &= \sum\limits_{k=1}^{4n+1}\frac{1}{k} + \sum\limits_{k=1}^{2n}\frac{1}{2k} - 2\sum\limits_{k=1}^{2n}\frac{1}{2k} -\sum\limits_{k=1}^n\frac{1}{2k}\\ &= \sum\limits_{k=1}^{4n+1}\frac{1}{k} + \frac{1}{2}\sum\limits_{k=1}^{2n}\frac{1}{k} - 2\sum\limits_{k=1}^{2n}\frac{1}{2k} -\sum\limits_{k=1}^n\frac{1}{2k}\\ &= \sum\limits_{k=1}^{4n+1}\frac 1k-2\sum\limits_{k=1}^{2n}\frac 1{2k}+\frac 12\sum\limits_{k=1}^{2n}\frac 1k-\sum\limits_{k=1}^n\frac 1{2k}. \end{align*}

Now if you write out each of these four sums, you get $$\underbrace{\left[\frac{1}{1} + \dots + \frac{1}{4n + 1}\right]}_{\textrm{all denominators}} - 2\underbrace{\left[\frac{1}{2} + \dots + \frac{1}{4n}\right]}_{\textrm{even denominators}} + \frac{1}{2}\underbrace{\left[\frac{1}{1} + \dots + \frac{1}{2n}\right]}_{\textrm{all denominators}} - \underbrace{\left[\frac{1}{2} + \dots + \frac{1}{2n}\right]}_{\textrm{even denominators}},$$ which can be viewed as $$\underbrace{\left[\frac{1}{1} + \dots + \frac{1}{4n + 1}\right]}_{\textrm{all denominators}} - 2\underbrace{\left[\frac{1}{2} + \dots + \frac{1}{4n}\right]}_{\textrm{even denominators}} + \frac{1}{2}\left(\underbrace{\left[\frac{1}{1} + \dots + \frac{1}{2n}\right]}_{\textrm{all denominators}} - 2\cdot\underbrace{\left[\frac{1}{2} + \dots + \frac{1}{2n}\right]}_{\textrm{even denominators}}\right).$$

Now you have a $1/k$ for each $1\leq k\leq 4n + 1$ in the first sum, and in the second sum you take away $2/k$ for each even $k$ between $1$ and $4n + 1$. This means that if you combine these, you wind up with $1/k$ for $1\leq k\leq 4n + 1$ when $k$ is even, and $-1/k$ for $1\leq k\leq 4n + 1$ when $k$ is odd. The same logic applies to the last two sums, and you wind up with the desired $$\sum\limits_{k=1}^{4n+1}\frac {(-1)^{k+1}}k+\frac 12\sum\limits_{k=1}^{2n}\frac {(-1)^{k+1}}k.$$

(P.S. you have an extra $k$ floating around in your problem statement - it should be $\sum\limits_{k=1}^{4n+1}\frac {(-1)^{k+1}}k+\frac 12\sum\limits_{k=1}^{2n}\frac {(-1)^{k+1}}k$, not $\sum\limits_{k=1}^{4n+1}\frac {(-1)^{k+1}}k+\frac 12\sum\limits_{k=1}^{2n}\frac {(-1)^{k+1}}k k$.)

What was Ramanujan's thinking when he split$$-\frac 12\sum\limits_{k=1}^{2n}\frac 1{k}=-2\sum\limits_{k=1}^{2n}\frac 1{2k}+\frac 12\sum\limits_{k=1}^{2n}\frac 1k$$

Again, I can't explain Ramanujan's personal thinking. He was extremely clever, and probably saw all the calculations performed above at a glance, and recognized that splitting up the sum in that way would let him combine terms into a pleasing/useful expression.

• I'm sorry, but I am still having trouble with this. For example, I'm not sure how the first equality is true.\begin{align*} \sum_{k=1}^{3n+1}\left(\frac 1{n+k}\right)-\sum_{k=1}^n\left(\frac 1{2n+2k}\right) &= \left[\sum_{k=1}^{n}\left(\frac 1{n+k}\right) + \sum_{k=n+1}^{3n+1}\left(\frac 1{n+k}\right)\right]-\sum_{k=1}^n\left(\frac 1{2n+2k}\right)\end{align*} – Crescendo Jun 17 '17 at 2:21
• In that step, I've just split up the sum into two sums. Instead of summing over all the $k$ from $1$ to $3n + 1$ at once, you can sum the $k$ from $1$ to $n$, and then add that to the sum of $k$ from $n + 1$ to $3n + 1$. Think: $(a + b + c + d) = (a + b) + (c + d)$. – Stahl Jun 17 '17 at 2:23
• Ah, okay. Thank you! I see now. The colours helped a lot! :) – Crescendo Jun 18 '17 at 22:06

In (1.1) you have $$\frac{1}{2n+1}+\frac{1}{2n+3}+\cdots+\frac1{4n+1}$$ (odd denominators) which equals $$\frac{1}{2n+1}+\frac{1}{2n+2}+\cdots+\frac1{4n+1}$$ (all denominators) minus $$\frac1{2n+2}+\frac1{2n+4}+\cdots+\frac1{4n}$$ (even denominators).