I was curious the other day about a limit that I could not figure out the answer to. The limit went somewhat like this: $$\lim_{N \to \infty}\sum_{k=1}^{N}( \frac{ln(x)}{k^2}+3^k)$$ My question is more about how to find the limit of this summation then to actually solve this summation. I know this exact summation of this limit is infinity but I want to know how to find the limit for other sums that don't go to infinity. And if this question is unclear be free to ask me or edit the question because I am new to this community don't want to have questions that are not good.
Sorry for editing the problem to much I was just trying to make it more clear but made it more complicated. I restored the actual problem so the answers would make sense.
Thanks to anyone that answers my problem!
$$\sum_{k=1}^N3^k=\frac{3(1-3^N)}{1-3}=\frac{3}{2}(3^N-1)$$ $$\sum_{k=1}^Nln(\frac{1}{k^2})=ln(\prod_{k=1}^N\frac{1}{k^2})=ln(\frac{1}{(N!)^2})=-2ln(N!)$$ $$\lim_{N\rightarrow\infty}\sum_{k=1}^N(3^k+ln(\frac{1}{k^2}))=\lim_{N\rightarrow\infty}(\frac{3^{N+1}}{2}-2ln(N!)-\frac{2}{3})$$ Also, $$2ln(N!)<2ln(N^N)=2NlnN<2N^2$$ for all $N>=1$
Meanwhile, $\frac{3^{N+1}}{2}$ is exponential and $2N^2$ is quadratic, so $$\lim_{N\rightarrow\infty}(\frac{3^{N+1}}{2}-2N^2)=\infty$$ Thus, $$\lim_{N\rightarrow\infty}\sum_{k=1}^N(3^k+ln(\frac{1}{k^2}))=\lim_{N\rightarrow\infty}(\frac{3^{N+1}}{2}-2ln(N!)-\frac{2}{3})=\infty$$