$f(x) = \Sigma_{n=1}^{\infty}b_nsin(nx) \ \forall x \in R$, where $R$ is the set of real numbers.
$b_n = \int_0^\pi 2e^{-4y}sin(ny)dy$
Find $f(\pi)$.
I think the answer is $0$, since $sin(n\pi) = 0\ \forall n \in N $ where $N$ is natural numbers. So if we add $0$ how many times as we want, we'll get $0$ only. So is my answer correct?
You are correct.
For all $n \in \Bbb{Z}$ we have $b_n \in \Bbb{R}$ and $\sin(n\pi) = 0$ hence $b_n \sin(n\pi) = 0$. Therefore the series representation for $f(\pi)$ is \begin{align*} f(\pi) &= \sum_{n=1}^{\infty} b_n \sin(n\pi) \\ &= \sum_{n=1}^{\infty} 0 \\ &= 0 \end{align*} In effect, the sequence of partial sums of the series $\displaystyle \sum_{n=1}^{\infty} 0$ is the constant sequence made up of $0$s hence the value of the series is $0$ : \begin{align*} \sum_{n=1}^{\infty} 0 &= \lim_{m \to \infty} \sum_{n=1}^{m} 0 && \text{by definition}\\ &= \lim_{m \to \infty} 0 && \text{finite sums of $0$s are $0$}\\ &= 0 \end{align*}