A relation involving summation and integration Let, $\displaystyle C(x)=\sum_{n\le x}c_n$, where $\{c_n\}$ is a sequence of complex numbers ; and let $f(t)$ be a continuously differentiable function such that $\displaystyle \lim_{Y\to \infty}C(Y)f(Y)=0$ and $\displaystyle \int_1^{\infty}C(t)f'(t)\,dt<\infty$. Then prove that $$\sum_{n=1}^{\infty}c_nf(n)+\int_1^{\infty}C(t)f'(t)\,dt=0.$$
$C(t)$ is a discrete sum, that's why I can't integrate by parts. So, in which way I've to proceed?
Any idea to prove this?
 A: Note that for $n \in \mathbb{N}$ and $t \in (n,n+1)$ we have $C(t) = \sum_{k=1}^n c_k$ , so the integral over such an interval is easy:
$$ \int \limits_n^{n+1} C(t) f'(t) \, \mathrm{d} x = \sum \limits_{k=1}^n c_k \int \limits_n^{n+1} f'(t) \, \mathrm{d} x = [f(n+1) - f(n)] \sum \limits_{k=1}^n c_k \, . $$
Now we can take $N \in \mathbb{N}$ and sum both sides to find
\begin{align}
\int \limits_1^{N+1} C(t) f'(t) \, \mathrm{d} x &= \sum \limits_{n=1}^N \int \limits_n^{n+1} C(t) f'(t) \, \mathrm{d} x = \sum \limits_{n=1}^N [f(n+1) - f(n)] \sum \limits_{k=1}^n c_k \\
&= f(N+1) C(N) + \sum \limits_{n=1}^{N-1} f(n+1) \sum \limits_{k=1}^n c_k - f(1) c_1- \sum \limits_{n=2}^N f(n) \sum \limits_{k=1}^n c_n  \, .
\end{align}
Shifting the index in the first sum reveals that most of the terms cancel:
\begin{align}
\int \limits_1^{N+1} C(t) f'(t) \, \mathrm{d} x &= f(N+1) C(N) - f(1) c_1 + \sum \limits_{n=2}^{N} f(n) \left[\sum \limits_{k=1}^{n-1} c_k - \sum \limits_{k=1}^n c_n \right] \\
&= f(N+1) C(N) - \sum \limits_{n=1}^N f(n) c_n \, .
\end{align}
Letting $N \to \infty$ concludes the proof.
A: For $0 < a < 1 < b$, we have $C(a) = 0$ and thus obtain the following identity as Riemann-Stieltjes integral.
$$ C(b)f(b) = \Big[ C(x)f(x) \Big]_{a}^{b} = \int_{a}^{b} d(C(x)f(x)) = \int_{a}^{b} C(x)f'(x) \, dx + \int_{a}^{b} f(x) \, dC(x). $$
We analyze the right-hand side.


*

*Since $C(x) = 0$ for $x < 1$, the first integral reduces to $\int_{a}^{b} C(x)f'(x) \, dx = \int_{1}^{b} C(x)f'(x) \, dx$.

*$\int_{a}^{b} f(x) \, dC(x) = \sum_{n=1}^{\lfloor b \rfloor} f(n) [C(n^+) - C(n^-)] = \sum_{n=1}^{\lfloor b \rfloor} f(n)c_n $.
Combining altogether, we obtain
$$ C(b)f(b) = \int_{1}^{b} C(x)f'(x) \, dx + \sum_{n=1}^{\lfloor b \rfloor} f(n)c_n. $$
In particular, $ \sum_{n=1}^{N} f(n)c_n = C(N)f(N) - \int_{1}^{N} C(x)f'(x) \, dx $ converges to $-\int_{1}^{\infty} C(x)f'(x) \, dx$ as $N\to\infty$ by the assumption and the desired identity holds.
