The analytic extension of the sum of the first $n$ reciprocals is given as
$$\sum_{k=1}^n\frac1k=\int_0^1\frac{x^n-1}{x-1}dx$$
I am wondering if $\frac1{n+1}+\sum_{k=1}^n\frac1k=\sum_{k=1}^{n+1}\frac1k$.
My attempt at checking
$$\sum_{k=1}^{1/2}\frac1k=\int_0^1\frac{\sqrt{x}-1}{x-1}dx$$
$$=\int_0^1\frac1{\sqrt{x}+1}dx$$
$$\int_0^1\frac1{\sqrt{x}+1}dx=\int_0^1\sum_{i=0}^\infty(-\sqrt{x})^idx\tag{Binomial Expansion}$$
$$=1+\sum_{i=1}^\infty(-1)^{i+1}\frac{2}{i}$$
$$=1+2\eta(1)$$
Where $\eta(n)$ is the Dirichlet eta function.
$$\sum_{k=1}^{1/2}\frac1k=1+2\eta(1)=1+2\ln(2)$$
I was wondering if I made any mistakes along the way and if it is ok for me to say that is the result.
Second problem:
$$\sum_{k=1}^{3/2}\frac1k=\int_0^1\frac{x^{3/2}-1}{x-1}dx$$
$$=\int_0^1\frac{(\sqrt{x}-1)(x+\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}dx$$
$$=\int_0^1\frac{x+\sqrt{x}+1}{\sqrt{x}+1}dx$$
$$=\int_0^1\sqrt{x}+\frac1{\sqrt{x}+1}dx$$
$$=\frac23+(1+2\ln(2))$$
I pulled that result from the previous problem.
Since this agrees with $\frac1{n+1}+\sum_{k=1}^n\frac1k=\sum_{k=1}^{n+1}\frac1k$, is it possible to prove that this will hold true for any $n$?
You can just try to calculate it directly: For $\alpha\geq 1$, we need to try the definition (analytical extension) of $\sum_{k=1}^\alpha\frac{1}{k}$ "appear" in the definition of $\sum_{k=1}^{\alpha+1}\frac{1}{k}$: \begin{align*} \sum_{k=1}^{\alpha+1}\frac{1}{k}&=\int_0^1\frac{x^{\alpha+1}-1}{x-1}dx=\int_0^1\frac{x^{\alpha+1}-x^\alpha+x^\alpha-1}{x-1}dx\\ &=\int_0^1x^\alpha+\int_0^1\frac{x^\alpha-1}{x-1}dx\\ &=\frac{1}{\alpha+1}+\sum_{k=1}^\alpha\frac{1}{k} \end{align*}
Note that these are almost the same calculations as you did in the second problem.