I am trying to find if there is a way of turning the following sum into an integral: $$\lim_{x\to\infty} \sum_{k=1}^{x}\operatorname{tr}\left(\frac{C^{k}}{x^2+x}\right)$$ I have looked into Riemann sums however I wasn't able to do it with this sum, however this could well be I didn't understand them fully.
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This is a geometric series. Assuming that $\|C\|<1$, $$\begin{split} \lim_{x\rightarrow+\infty}\sum_{k=1}^{x}\operatorname{tr}\left(\frac{C^{k}}{x^2+x}\right) &= \lim_{x\rightarrow+\infty}\frac 1 {x^2 + x}\operatorname{tr}\left(\sum_{k=1}^{x}C^{k}\right)\\ &= \lim_{x\rightarrow+\infty}\frac 1 {x^2 + x}\operatorname{tr}\left(C(I-C^x)(I-C)^{-1}\right)\\ &= \operatorname{tr}\left(C(I-C)^{-1}\right)\lim_{x\rightarrow+\infty}\frac 1 {x^2+x}\\ &=0 \end{split}$$
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$\begingroup$ Then we can't say that the series converges. $\endgroup$ Commented Jan 3, 2021 at 21:29