Evaluating the limit $$\lim_{n \to \infty} \sum_{k=1}^{n} \left(1-\frac{1}{k+3}\right)$$ I encountered a doubt: I did this $$\lim_{n \to \infty} \sum_{k=1}^{n} \left(1-\frac{1}{k+3}\right)=\lim_{n \to \infty} \left(\sum_{k=1}^{n} 1-\sum_{k=1}^{n}\frac{1}{k+3}\right)=\lim_{n \to \infty} \left(n-\sum_{k=1}^{n}\frac{1}{k+3}\right)$$ Letting $r:=k+3$ in the second sum I get $$\lim_{n \to \infty} \left(n-\sum_{k=1}^{n} \frac{1}{k+3} \right)=\lim_{n \to \infty} \left(n-\sum_{r=4}^{n+3} \frac{1}{r} \right)=\lim_{n \to \infty} \left[n-\left(\sum_{r=1}^{n}\frac{1}{r}-1-\frac{1}{2}-\frac{1}{3}+\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}\right)\right]=$$ $$=\lim_{n \to \infty} \left[n-\left(\frac{1}{n}\sum_{r=1}^{n}\frac{1}{\frac{r}{n}}-1-\frac{1}{2}-\frac{1}{3}+\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}\right)\right]$$ Now the doubt is the following: I know that $$\lim_{n\to \infty} \frac{1}{\ln n} \sum_{k=1}^n \frac{1}{k}=1$$ So the latter sum is asymptotic to $\ln n$, but I know too that $$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \frac{1}{\frac{k}{n}}=\int_0^1 \frac{1}{x} dx$$ So I would conclude that $$\lim_{n \to \infty} \left[n-\left(\frac{1}{n}\sum_{r=1}^{n}\frac{1}{\frac{r}{n}}-1-\frac{1}{2}-\frac{1}{3}+\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}\right)\right]=\infty$$ Because the infinite from the improper integral $\int_0^1 \frac{1}{x} dx$ goes as a logarithm, so the indeterminate form $\infty-\infty$ can be determined dividing by $n$; but I'm not sure this is true because
1) I should let $n \to \infty$ to get the definition of integral, and there is that $n$ too that would go to infinity too; also, the improper integral should be something like this $\lim_{a->0^+} \int_a^1 \frac{1}{x} dx$ and so I would get a double limit and I don't know how to approach it.
2) I'm sure that Riemann sums works the same for improper integrals.
Can someone tell me if this approach is correct and if it can be improved and made rigorous? Thanks.
A necessary condition for the convergence of $\sum_{k=1}^{\infty}a_k$ is $\lim_{k\to\infty}a_k=0$. Can you see that this is not satisfied here? Since the series is divergent and its terms are positive, the limit of the partial sums is $\infty$.
We can also work with comparisons. Since $\sum_{k=1}^n 1/k\leq\ln(n)+1$, we have $$\sum_{k=1}^n\left(1-\frac{1}{k+3}\right)\geq n-\ln(n+3)\to\infty\;\text{as}\;n\to\infty $$ so that again implies divergence.