Approximation to a partial sum

Question

Just like $$\sum_{k=1}^{n}\frac1k$$ can be approximated by log(n), is there a similar approximation for the sum $$\sum_{k=1}^{n}\frac1{k^2}$$

Answer 1

From

$$x-1<\lfloor x\rfloor\le x$$ you draw

$$\frac1{x^d}\le\frac1{\lfloor x\rfloor^d}<\frac1{(x-1)^d}.$$

Then by integration from $2$ to $n+1$,

$$\int_{x=2}^{n+1}\frac{dx}{x^d}\le\sum_{k=2}^n\frac1{k^d}<\int_{x=2}^{n+1}\frac{dx}{(x-1)^d}.$$

So

$$\ln\frac{n+1}2\le\sum_{k=2}^n\frac1{k}<\ln n$$ and

$$\frac12-\frac1{n+1}\le\sum_{k=2}^n\frac1{k^2}<1-\frac1{n}.$$

(Mind the initial index $2$.)

You can get tighter approximations by starting from larger lower bounds and adding the first terms separately.

Answer 2

The answer is that, using $\;\psi_1(x)\;$ the trigamma function, we have $\;\sum_{k=1}^n \frac1{k^2} = \psi_1(1)-\psi_1(n+1). $ It is known that $\;\psi_1(x) = \frac1x + \frac1{2x^2} + \cdots = \sum_{k=0}^\infty \frac{B_k}{x^{k+1}}$ asymptotically using Bernoulli numbers.