Fourier coefficients of product of two periodic signals question:
If $x(t)$ and $y(t)$ are two periodic signals(both with period T) with Fourier coefficients $c_{n}$ and $d_{n}$ respectively then, Fourier coefficient of $z(t)=x(t)\cdot y(t)$ is:

(a) $\dfrac{c_{n}d_{n}}{T}$
(b) $T c_{n}d_{n}$
(c) $c_{n}d_{n}$
(d) $c_{n}*d_{n}$

my attempt :

let, $x(t)=\sum_{-\infty}^{\infty} c_{n}\cdot e^{jn\omega_{0}t}$ and $y(t)=\sum_{-\infty}^{\infty} d_{n}\cdot e^{jn\omega_{0}t}$.

Then, fourier cofficient of $z(t)$ will be equal to

$$\begin{align}
\dfrac{1}{T}\displaystyle\int_{0}^{T} z(t) e^{-jn\omega_{0} t}dt
&amp;=\dfrac{1}{T}\displaystyle\int_{t=0}^{T} x(t)\cdot y(t) e^{-jn\omega_{0} t}dt \\
&amp;=\dfrac{1}{T}\displaystyle\int_{t=0}^{T} \left(\sum_{l=-\infty}^{\infty} c_{l}\cdot e^{jl\omega_{0}t}\sum_{m=-\infty}^{\infty} d_{m} \cdot e^{jm\omega_{0}t} \right)e^{-jn\omega_{0} t}dt\\
&amp;= \dfrac{1}{T}\displaystyle\int_{t=0}^{T}\left(\sum_{l=-\infty}^{\infty}\sum_{m=-\infty}^{\infty}c_{l}d_{m}\cdot e^{j(l+m)\omega_{0}t}\right)e^{-jn\omega_{0}t}dt
\end{align}$$

[in above $l$, $m$ are dummy variables]

after changing order of integration and summation i got this:

$$\begin{align}
\displaystyle\sum_{l=-\infty}^{\infty}\sum_{m=-\infty}^{\infty}c_{l}d_{m}\left[\dfrac{1}{T}\int_{t=0}^{T}e^{-j(n-l-m)\omega_{0} t}dt\right]=\displaystyle\sum_{l=-\infty}^{\infty}\sum_{m=-\infty}^{\infty}c_{l}d_{m}\cdot\delta[n-l-m]\end{align}$$

where

$$\delta[n] \triangleq \begin{cases}
1 \qquad &amp; \text{for } n = 0 \\
0 \qquad &amp; \text{for } n \in \mathbb{Z}, n \ne 0 \\
\text{undefined} \quad &amp; \text{otherwise} \\
\end{cases}$$
is the Kronecker delta.

But from here on I'm unable to proceed,

i know answer is option (d)i.e, $c_{n}*d_{n}$ but i am interested in proving the result formally.
You did everything right, and you're almost there. Your last equation becomes

$$\sum_lc_l\sum_md_m\delta[n-l-m]\tag{1}$$

Due to the Kronecker delta the sum over $m$ reduces to a single element:

$$\sum_md_m\delta[n-l-m]=d_{n-l}\tag{2}$$

because $\delta[n-l-m]$ is non-zero for $m=n-l$ and zero otherwise. Plugging $(2)$ into $(1)$ gives the desired result:

$$\sum_lc_ld_{n-l}$$