Comparing integral and series of $1/(x^a)$ The problem is to $$\sum^N_{n=2}\frac{1}{n^a}\leq\int^N_1\frac{1}{x^a}$$ and to use this to prove the convergence of the series for $a>1$.
So, I believe I have the second part down. Namely evaluating the integral for the cases


*

*$$a>1\Rightarrow\int^N_1\frac{1}{x^a}=\frac{1}{1-a}$$

*$$a=1\Rightarrow\int^N_1\frac{1}{x^a}=\infty$$

*$$0<a<1\Rightarrow\int^N_1\frac{1}{x^a}=\infty$$
and so using the [unproven fact] that $\sum^N_{n=2}\frac{1}{n^a}\leq\int^N_1\frac{1}{x^a}$, the latter part follows from the Comparison Test. Still, I am having a hard time proving the former part of this question. 

 A: First notice that if $n\in\mathbb N$, since $1/x^a\gt 1/(n+1)^a$ for all $x\in (n,n+1)$, we have that
$$\int_n^{n+1}\frac{dx}{x^a}\gt \int_n^{n+1}\frac{dx}{(n+1)^a}=\frac{1}{(n+1)^a}$$
This implies that
$$\begin{align}\
\int_1^N \frac{dx}{x^a} &=\int_1^2 \frac{dx}{x^a}+\int_2^3 \frac{dx}{x^a}+...+\int_{N-1}^N \frac{dx}{x^a}\\
&> \frac{1}{2^a}+\frac{1}{3^a}+...+\frac{1}{N^a}\\
&=\sum_{n=2}^N \frac{1}{n^2}
\end{align}$$
which proves the desired inequality:
$$\int_1^N \frac{dx}{x^a}\gt \sum_{n=2}^N \frac{1}{n^2}$$
A: Make a sketch for $\frac1{x^a}$ and $\frac1{(x-1)^a}$ 

then we have
$$\sum^N_{n=2}\frac{1}{n^a}\leq\int^{N+1}_2\frac{1}{(x-1)^a}=\int^N_1\frac{1}{x^a}$$
A: We show that $${1\over (k+1)^a}\le \int_{k}^{k+1} {1\over x^a}dx$$and then by summing up the sided from $k=1$ to $k=N-1$ we are done. From the other side we know $${1\over (k+1)^a}=\int_k^{k+1} {dx\over (\lfloor x\rfloor +1)^a}$$by the definition of floor function. Also $$\lfloor x\rfloor\le x<\lfloor x\rfloor+1$$therefore for $x\ge 1$ and $a>1$ we obtain  $$(\lfloor x\rfloor+1)^a>x^a$$or equivalently $${1\over (\lfloor x\rfloor+1)^a}<{1\over x^a}$$by integrating the both sides, we prove the integral inequality first, hence the general problem. 
