Why is it that $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$ is not less than $\int_1^\infty \frac{dx}{x^2} = 1$?

Question

So according to Euler's proof of the Basel problem,

$$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6},$$

But only for $n \in \mathbb{Z}$.

But if $n$ was a positive real and $n \geqslant 1$, then would the sum $S$ be equal to,

$$\int_1^\infty \frac{dx}{x^2}?$$

If so then after solving the integral via power rule we get $S=1$.

But how can this be since when taking reals, we take integral values as well as new values. So by logic shouldn't, $S>\dfrac{\pi^2}{6}$ but $1$ isn't. Where am I going wrong pls explain.

Answer 1

Note that

$$S=\int_1^1 \frac{dx}{x^2}=0\quad \neq \quad \sum_{n=1}^1 \frac{1}{n^2}=1$$

To see more in general why inquality doesn't hold you should consider the function $\frac1{x^2}$ and compare with $\frac1{n^2}:=\frac{1}{\lfloor x\rfloor^2}\ge\frac{1}{\lfloor x^2\rfloor}$ to observe that

$$\frac1{x^2} \le \frac{1}{\lfloor x^2\rfloor}\le\frac{1}{\lfloor x\rfloor^2}\implies \int_1^\infty\frac{dx}{x^2}\le \int_1^\infty\frac{dx}{\lfloor x\rfloor^2}= \sum_{n=1}^\infty \frac{1}{n^2}$$

Plot for latter inequality

Answer 2

Why should the integral be equal to the sum? To see the sum must be larger, note that $$\int_1^\infty\frac{dx}{x^2}=\sum_{n=1}^\infty \int_n^{n+1}\frac{dx}{x^2}<\sum_{n=1}^\infty\frac1{n^2}$$ as $x\mapsto 1/x^2$ is a decreasing function.

Answer 3

$$\sum_{k=1}^{\infty}\frac1{k^2}-\int_1^{\infty}\frac{dx}{x^2}=\sum_{k=1}^{\infty}\left\{\frac1{k^2}-\int_k^{k+1}\frac{dx}{x^2}\right\}>0$$ Because in each interval of integration, $$\frac1{k^2}>\frac1{x^2}$$

Answer 4

In the following picture, the pink area is the left-hand part of the integral $\displaystyle \int_{x=1}^\infty \frac1x \, dx$ while the green and pink areas together are left-hand part of the sum $\displaystyle \sum_{n=1}^\infty \frac1{n^2}$

The green area is the difference, which is clearly positive and is in fact $\dfrac{\pi^2}{6} -1\approx 0.6449$

Answer 5

Your logic doesn't hold because the integrand is a decreasing function so that its average value in a unit interval is lower than its initial value.

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As the average value is larger than the final value, the following bracketing is guaranteed:

$$\dfrac{\pi^2}6-1<\int_1^\infty\frac{dx}{x^2}<\dfrac{\pi^2}6.$$

$\color{lightgreen}{\text{initial}},\color{blue}{\text{average}},\color{magenta}{\text{final}}$

Answer 6

Each term of the sum is greater than the corresponding part of the integral. Therefore, the sum is greater than the integral. $$ \begin{align} \sum_{k=1}^\infty\frac1{k^2}-\int_1^\infty\frac{\mathrm{d}x}{x^2} &=\sum_{k=1}^\infty\left(\frac1{k^2}-\int_k^{k+1}\frac{\mathrm{d}x}{x^2}\right)\\ &=\sum_{k=1}^\infty\left(\frac1{k^2}-\frac1{k(k+1)}\right)\\ &=\sum_{k=1}^\infty\frac1{k^2(k+1)}\\[6pt] &\ge\frac12\quad\text{(the first term)} \end{align} $$

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The sum is the area of the rectangles $\left(\frac1{k^2}\right)$ and the integral is the area below the curve, $\left(\frac1{k(k+1)}\right)$. The difference is the area of the rectangles above the curve $\left(\frac1{k^2(k+1)}\right)$.

In the image above, it can be seen that $$ \sum_{k=1}^\infty f(k)=\int_1^\infty f(\lfloor x\rfloor)\,\mathrm{d}x $$ If $f$ is a decreasing function, then $f(x)\le f(\lfloor x\rfloor)$ because $x\ge\lfloor x\rfloor$, and therefore, $$ \begin{align} \int_1^\infty f(x)\,\mathrm{d}x &\le\int_1^\infty f(\lfloor x\rfloor)\,\mathrm{d}x\\ &=\sum_{k=1}^\infty f(k) \end{align} $$

Answer 7

If I guess correctly you want to ask why the sum of $1/n^2$ is greater than integral of $1/x^2$ whereas logically it should be the other way round. Your argument is that the integral actually tries to sum all the values for $1/x^2$ and not just the ones where $x$ is integer. Sorry!! Thus is wrong. An integral is not summing the values of a function. Rather it is summing product of values of function and the length of subintervals based on which function values are selected. Remember the Riemann sum is not $\sum f(x_i) $ but rather $\sum f(x_i) (x_i-x_{i-1})$ and it tends to the value of integral when the subinterval length $x_i-x_{i-1}$ tends to zero.

On the other hand the sum $\sum 1/n^2$ can be easily seen to be some kind of a Riemann sum where interval length is fixed as $1$ so obviously this has to be greater than integral of $1/x^2$ (full justification of this is already available in other answers).

To be precise the idea of an integral is not sum a continuous set of values but rather a limit of a very specific kind of sum. It is however possible to define sum of a continuous set of values (sum over an uncountable index set) but it turns out to be infinite unless each term is $0$.