 During his short life, Ramanujan independently compiled nearly 3,900 results (mostly identities and equations). Many were completely novel; his original and highly unconventional results, such as the Ramanujan prime, the Ramanujan theta function, partition formulae and mock theta functions, have opened entire new areas of work and inspired a vast amount of further research. Of his thousands of results, all but a dozen or two have now been proven correct. The Ramanujan Journal, a scientific journal, was established to publish work in all areas of mathematics influenced by Ramanujan, and his notebooks—containing summaries of his published and unpublished results—have been analysed and studied for decades since his death as a source of new mathematical ideas. As late as 2012, researchers continued to discover that mere comments in his writings about “simple properties” and “similar outputs” for certain findings were themselves profound and subtle number theory results that remained unsuspected until nearly a century after his death. He became one of the youngest Fellows of the Royal Society and only the second Indian member, and the first Indian to be elected a Fellow of Trinity College, Cambridge. Of his original letters, Hardy stated that a single look was enough to show they could have been written only by a mathematician of the highest calibre, comparing Ramanujan to mathematical geniuses such as Euler and Jacobi.

# Understanding these basic definition and statement~

• Let $\left(a_{k}\right)$ be a sequence of real numbers. We use the notation $s_{n}=\sum_{k=0}^{n} a_{k}$ to denote the $n$th partial sum of the infinite series $s_{\infty}=\sum_{k=0}^{\infty} a_{k} .$ If the sequence of partial sums $\left(s_{n}\right)$ converges to a real number $s$, we say that the series $\sum_{k} a_{k}$ is convergent and we write $s=\sum_{k=0}^{\infty} a_{k} . \mathrm{A}$ series that is not convergent is called divergent.
• An infinite series $\sum_{k=0}^{\infty} a_{k}$ is said to converge absolutely if $\sum_{k=0}^{\infty}\left|a_{k}\right|$ converges. If a series converges absolutely, then it converges. Furthermore, an absolutely convergent series converges to the same sum in whatever order the terms are taken.

If $\sum^{\infty} a_{k}$ converges but $\sum_{k}^{\infty}\left|a_{k}\right|$ diverges, then we say $\sum^{\infty} a_{k}$ converges conditionally. Any conditionally convergent series can be rearranged to obtain a series which converges to any given sum or diverges to $\infty$ or $-\infty .$

• Ratio Test: Given a series $\sum_{k=1}^{\infty} a_{k}$ with $a_{k} \neq 0$, if $a_{k}$ satisfies
$$\lim {k \rightarrow \infty}\left|\frac{a{k+1}}{a_{k}}\right|=r<1$$
then the series converges absolutely.
• Root Test: Let $\sum_{k=1}^{\infty} a_{k}$ be a series and
$$\alpha:=\lim {k \rightarrow \infty} \sqrt[k]{\left|a{k}\right|}$$
Then the following hold.
a) $\sum_{k=1}^{\infty} a_{k}$ converges absolutely if $\alpha<1$. b) $\sum_{k=1}^{\infty} a_{k}$ diverges if $\alpha>1$.
For $\alpha=1$ both convergence and divergence of $\sum_{k=1}^{\infty} a_{k}$ are possible.
• Cauchy Criterion for Series: The series $\sum^{\infty} a_{k}$ converges if and only if given $\varepsilon>0$, there exists $N \in \mathbb{N}$ such that whenever $n>m \geq N$ it follows that
$$\left|a_{m+1}+a_{m+2}+\cdots+a_{n}\right|<\varepsilon$$
• Comparison Test: Assume $\left(a_{k}\right)$ and $\left(b_{k}\right)$ are sequences satisfying $0 \leq a_{k} \leq b_{k}$ for all $k \in \mathbb{N}$.
a) If $\sum^{\infty} b_{k}$ converges, then $\sum^{\infty} a_{k}$ converges.
b) If $\sum_{k=1}^{\infty} a_{k}$ diverges, then $\sum_{k=1}^{\infty} b_{k}$ diverges.
• Geometric Series: A series is called geometric if it is of the form
$$\sum_{k=1}^{\infty} a r^{k}=a+a r+a r^{2}+a r^{3} \cdots$$
and
$$\sum_{k=1}^{\infty} a r^{k}=\frac{a}{1-r}$$
if and only if $|r|<1 .$ In case $r=1$ and $a \neq 0$, the series diverges.
• Alternating Series Test: Let $\left(a_{n}\right)$ be a sequence satisfying
a) $a_{1} \geq a_{2} \geq a_{3} \geq \cdots \geq a_{n} \geq a_{n+1} \geq \cdots$ and
b) $\left(a_{n}\right) \rightarrow 0$.
Then the alternating series $\sum_{n=1}^{\infty}(-1)^{n+1} a_{n}$ converges.
• Let $\sum_{k=1}^{\infty} a_{k}$ be a series. A rearrangement is the series $\sum_{k=1}^{\infty} a_{\sigma(k)}$ where $\sigma$ is a permutation of ${1,2,3, \ldots} .$ The summands of the rearrangement $\sum_{k=1}^{\infty} a_{\sigma(k)}$ are the same as those of the original series, but they occur in different order. If $\sigma$ is a permutation of $\mathbb{N}$ with $\sigma(k)=k$ for almost all $k \in \mathbb{N}$, then $\sum_{k=1}^{\infty} a_{k}$ and $\sum_{k=1}^{\infty} a_{\sigma(k)}$ have the same convergence behavior, and their values are equal if the series converge. For a permutation $\sigma(k) \neq k$ for infinitely many $k \in \mathbb{N}$, this may not be true.
• If $\sum_{k=1}^{\infty} a_{k}$ converges absolutely, then any rearrangement of this series converges to the same limit.

# Try the following problem to test yourself！

Problem 1.

Show that the Euler’s series $\sum_{k=1}^{\infty} \frac{1}{k^{2}}$ converges. Find the sum of the series.

Proof .

Let us use the substitution $u=\pi-x .$ Then we have
$$\int_{0}^{\pi} x e^{\sin (x)} d x=\int_{\pi}^{0}-(\pi-u) e^{\sin (\pi-u)} d u=\int_{0}^{\pi}(\pi-u) e^{\sin (u)} d u$$
which implies
$$\int_{0}^{\pi} x e^{\sin (x)} d x=\pi \int_{0}^{\pi} e^{\sin (u)} d u-\int_{0}^{\pi} u e^{\sin (u)} d u$$
Since
$$\int_{0}^{\pi} x e^{\sin (x)} d x=\int_{0}^{\pi} u e^{\sin (u)} d u$$
we get

Because the terms in the sum are all positive, the sequence of partial sums given by
$$s_{n}=1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^{2}}$$
is increasing. To find an upper bound for $s_{n}$, observe
\begin{aligned} s_{n} &=1+\frac{1}{2 \cdot 2}+\frac{1}{3 \cdot 3}+\cdots+\frac{1}{n \cdot n} \ &<1+\frac{1}{2 \cdot 1}+\frac{1}{3 \cdot 2}+\cdots+\frac{1}{n \cdot(n-1)} \ &=1+\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3 \cdot 2}\right)+\cdots+\left(\frac{1}{n-1}-\frac{1}{n}\right) \ &=1+1-\frac{1}{n} \ &<2 . \end{aligned}

Thus 2 is an upper bound for the sequence of partial sums, so by the Monotone Convergence Theorem, $\sum_{k=1}^{\infty} \frac{1}{k^{2}}$ converges to a limit less than 2 . Next, we claim that $\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}$. This can be shown by using the well-known clever trick of evaluating the double integral
$$I=\int_{0}^{1} \int_{0}^{1} \frac{1}{1-x y} d x d y$$
and we evaluate $I$ in two different ways. First notice
$$\frac{1}{1-x y}=\sum_{k=0}^{\infty}(x y)^{k},$$
therefore
\begin{aligned} I &=\int_{0}^{1} \int_{0}^{1} \sum_{k=0}^{\infty}(x y)^{k} d x d y \ &=\sum_{k=0}^{\infty} \int_{0}^{1} \int_{0}^{1}(x y)^{k} d x d y \ &=\sum_{k=0}^{\infty}\left(\int_{0}^{1} x^{k} d x\right)\left(\int_{0}^{1} y^{k} d y\right) \ &=\sum_{k=0}^{\infty} \frac{1}{(k+1)^{2}} \ &=\sum_{k=1}^{\infty} \frac{1}{k^{2}} \end{aligned}

The second way to evaluate $I$ comes from a change of variables. Let
$$u=\frac{x+y}{2} \quad \text { and } \quad v=\frac{y-x}{2}$$
or equivalently
$$x=u-v \quad \text { and } \quad y=u+v$$
Given this transformation,
$$\frac{1}{1-x y}=\frac{1}{1-\left(u^{2}+v^{2}\right)}$$
and using the change of variables formula we obtain
$$I=\iint f(x, y) d x d y=\iint f(x(u, v), y(u, v))\left|\frac{d(x, y)}{d(u, v)}\right| d u d v$$
where
$$\left|\frac{d(x, y)}{d(u, v)}\right|=2$$

Since the function to be integrated and the domain in the $u v$-plane are symmetric with respect to the $u$-axis, we can split the integral into two parts as such,
\begin{aligned} I &=4 \int_{0}^{1 / 2}\left(\int_{0}^{u} \frac{d v}{1-u^{2}+v^{2}}\right) d u+4 \int_{1 / 2}^{1}\left(\int_{0}^{1-u} \frac{d v}{1-u^{2}+v^{2}}\right) d u \ &=4 \int_{0}^{1 / 2} \frac{1}{\sqrt{1-u^{2}}} \arctan \left(\frac{u}{\sqrt{1-u^{2}}}\right) d u+4 \int_{1 / 2}^{1} \frac{1}{\sqrt{1-u^{2}}} \arctan \left(\frac{1-u}{\sqrt{1-u^{2}}}\right) d u . \end{aligned}
Now, observe that if we set
$$k(u)=\arctan \left(\frac{u}{\sqrt{1-u^{2}}}\right) \text { and } h(u)=\arctan \left(\frac{1-u}{\sqrt{1-u^{2}}}\right),$$
then we obtain the derivatives
$$k^{\prime}(u)=\frac{1}{\sqrt{u^{2}}} \text { and } h^{\prime}(u)=-\frac{1}{2} \frac{1-u}{\sqrt{1-u^{2}}} \text {. }$$
This yields
\begin{aligned} I &=4 \int_{0}^{1 / 2} k^{\prime}(u) k(u) d u+4 \int_{1 / 2}^{1}-2 h^{\prime}(u) h(u) d u \ &=\left.2(k(u))^{2}\right|{0} ^{1 / 2}-\left.4(h(u))^{2}\right|{1 / 2} ^{1} \ &=2(k(1 / 2))^{2}-2(k(0))^{2}–4(h(1))^{2}+4(h(1 / 2))^{2} \ &=2\left(\frac{\pi}{6}\right)^{2}-0+0+4\left(\frac{\pi}{6}\right)^{2} \ &=\left(\frac{\pi}{6}\right)^{2} . \end{aligned}

Problem 2.

Let $T:(C[0,1],|\cdot|) \rightarrow(C[0,1],|\cdot|)$ be a function defined as
$$T f(x)=\int_{0}^{x} f(t) d t$$
where by $C[0,1]$ we mean the vector space of all continuous real-valued functions defined on $[0,1]$ and $|f|=\sup _ Discuss the convergence or divergence of$\sum x_{n}$where $$x_{n}=\frac{1 \cdot 3 \cdots(2 n-1)}{2 \cdot 4 \cdots(2 n)}$$ Proof . Note that $$x_{n}=\frac{(2 n) !}{\left(2^{n} n !\right)^{2}}=\frac{(2 n) !}{2^{2 n}(n !)^{2}}$$ Let us first remark that$\lim {n \rightarrow \infty} x{n}=0 .$Indeed, notice that$0<x_{n}<1$for any$n \geq 1$. Since$x_{n+1}=\frac{2 n+1}{2 n+2} x_{n}$, we have$x_{n+1}<x_{n}$for any$n \geq 1$, which implies that$\left{x_{n}\right}$is decreasing. So$\lim {n \rightarrow \infty} x{n}=l$exists. Define $$y_{n}=\frac{2 \cdot 4 \cdots(2 n)}{3 \cdot 5 \cdots(2 n+1)}$$ Since$4 n^{2}-1<4 n^{2}$we get$\frac{2 n-1}{2 n}<\frac{2 n}{2 n+1}$, for any$n \geq 1 .$This obviously implies$x_{n}<y_{n}$for any$n \geq 1 .$Since$x_{n} y_{n}=\frac{1}{2 n+1}$we deduce that$x_{n}^{2}<x_{n} y_{n}=\frac{1}{2 n+1}$for any$n \geq 1 .$Hence$\lim {n \rightarrow \infty} x{n}^{2}=0$which implies$\lim {n \rightarrow \infty} x{n}=0 .$This conclusion may suggest that the series$\sum x_{n}$is convergent. But since $$\frac{x_{n+1}}{x_{n}}=\frac{2 n+1}{2 n+2}$$ we have $$n\left(1-\frac{x_{n+1}}{x_{n}}\right)=\frac{n}{2 n+2}$$ Hence$\lim {n \rightarrow \infty} n\left(1-\frac{x{n+1}}{x_{n}}\right)=\frac{1}{2}<1$, then$\sum x_{n}$is divergent based on the previous problem. Note that the ratio test does not help since$\lim {n \rightarrow \infty} \frac{x{n+1}}{x_{n}}=1 .$In fact, the root test will also be not conclusive. Problem 3. Given two series$\sum x_{n}$and$\sum y_{n}$, define$z_{n}=\sum_{k=0}^{n} x_{k} y_{n-k} .$Suppose that$\sum x_{n}$and$\sum y_{n}$are absolutely convergent. Show that$\sum z_{n}$is absolutely convergent, and $$\sum z_{n}=\sum x_{n} \cdot \sum y_{n} .$$ Proof . We show that the series $$x_{0} y_{0}+x_{0} y_{1}+x_{1} y_{1}+x_{1} y_{0}+x_{0} y_{2}+x_{1} y_{2}+x_{2} y_{2}+x_{2} y_{1}+x_{2} y_{0}+\cdots$$ converges absolutely. In fact, the sum of the first$(n+1)^{2}$terms of the series$(8.4)\left|x_{0} y_{0}\right|+\left|x_{0} y_{1}\right|+\left|x_{1} y_{1}\right|+\left|x_{1} y_{0}\right|+\left|x_{0} y_{2}\right|+\left|x_{1} y_{2}\right|+\left|x_{2} y_{2}\right|+\left|x_{2} y_{1}\right|+\left|x_{2} y_{0}\right|+\cdots$is$\sum_{k=0}^{n}\left|x_{k}\right| \cdot \sum_{k=0}^{n}\left|y_{k}\right|$, which converges to$\sum_{k=0}^{\infty}\left|x_{k}\right| \cdot \sum_{k=0}^{\infty}\left|y_{k}\right|$. Hence the sequence of partial sums of$(8.4)$has a convergent subsequence. But all the terms of the series are positive, so the sequence of all partial sums is increasing, and bounded above by$\sum_{k=0}^{\infty}\left|x_{k}\right| \cdot \sum_{k=0}^{\infty}\left|y_{k}\right|$, so it also converges, to the same limit. Thus (8.3) converges absolutely. The same argument as above, considering sums of the first$(n+1)^{2}$terms of the series$(8.3)$, shows that the sum of this series is$\sum x_{n} \cdot \sum y_{n}$But $$\sum z_{n}=x_{0} y_{0}+\left(x_{0} y_{1}+x_{1} y_{0}\right)+\left(x_{0} y_{2}+x_{1} y_{1}+x_{2} y_{0}\right)+\cdots$$ is a rearrangement of$(8.3)$, so by the Rearrangement Theorem it converges to the same limit. Note: You might like to think about the above proof by considering an “infinite matrix” in which the$(i, j)$term is$x_{i} y_{i}$:$\begin{array}{lllll}x_{0} y_{0} & x_{0} y_{1} & x_{0} y_{2} & x_{0} y_{3} & \cdots \ x_{1} y_{0} & x_{1} y_{1} & x_{1} y_{2} & x_{1} y_{3} & \cdots \ x_{2} u_{0} & x_{2} u_{1} & x_{2} u_{2} & x_{2} y_{2} & \cdots\end{array}$$$\cdots$$$0 \quad x_{3}\$
$$\begin{array}{r} \cdots \ \ddots \end{array}$$