Why is the LHS of this inequality a series uniform and absolute convergence?
It's about integral operators... let's see $f\in L^2(\mu)$, $\lambda_n$ a sequence of eigenvalues, $\varphi_n$ a sequence of eigenvectors, X a compact space and $M$ just a constant. I don't want to prove this inequality just need to deduct the convergence of first series. –
$$\sum_{n=p}^q\left|\lambda_n\left(\int_{X}f(y)\;\overline{\varphi_n}(y)\;d\mu(y) \right)\varphi_n(x)\right|\leq M\cdot \left({\sum_{n=p}^q\left|\int_{X}f(y)\;\overline{\varphi_n}(y)\;d\mu(y)\right|^2}\right)^{1/2}$$
It's useful to say I guess that $$\langle f,\varphi_n\rangle=\int_{X}f(y)\,\overline{\varphi_n}(y)\;d\mu(y)$$
A series $\sum_n a_n$ converges (be definition) if the sequence of partial sums $$ \left\{\sum_{n=1}^ma_n\right\}_m $$ converges. This sequence of partial sums converges if and only if it is Cauchy: that is, given $\varepsilon>0$ there exists $m_0$ such that whenever $q>p\geq m_0$, $$ \left|\sum_{n=1}^p a_n-\sum_{n=1}^q a_n\right|<\varepsilon. $$ Now note that $$ \sum_{n=1}^p a_n-\sum_{n=1}^q a_n=\sum_{n=p+1}^q a_n. $$ The inequality you have shows that the tails of the series on the left are bounded by the tails of the series on the right. The series on the right is convergent (Parseval).