Why is the inequality $\sum_{n=1}^{\infty} \frac{1}{n^2} \leq 1 + \int_1^{\infty} \frac{1}{x^2}$ true?

Question

$$\sum_{n=1}^{\infty} \frac{1}{n^2} \leq 1 + \int_1^{\infty} \frac{1}{x^2}dx$$

I'm having trouble figuring out why the inequality above is true. I understand the following inequality:

$$\int_1^{\infty} \frac{1}{x^2}dx \leq \sum_{n=1}^{\infty} \frac{1}{n^2}$$

It makes sense because the rectangles formed from the right side of the inequality have pieces that go over-top of the function like so:

So let's say I rewrite $\sum_{n=1}^{\infty} \frac{1}{n^2}$ as $1 + \sum_{n=2}^{\infty} \frac{1}{n^2}$ since they are equivalent.

Why is that less than $1 + \int_1^{\infty} \frac{1}{x^2}dx$?

If I picture it in my head it sounds like its saying that the first term is greater than all the little pieces that form above the curve in the picture above.

My claims come specifically from page 60 of this webpage from Dartmouth

Answer 1

A way to see that from the graph is as follows

$$\sum_{n=1}^{\infty} \frac{1}{n^2} \leq 1 + \int_2^{\infty} \overbrace{\frac{1}{(x-1)^2}}^{graph\, for\, \frac1{x^2}\,shifted \, by\, 1}dx= 1 + \int_1^{\infty} \frac{1}{x^2}dx$$

Answer 2

The right endpoint sums for the integral have the form:

$$ \sum_{n = 2}^\infty \frac{1}{n^2} $$ and we know: $$ \sum_{n = 2}^\infty \frac{1}{n^2} \leq \int_1^\infty \frac{1}{x^2} \leq \sum_{n = 1}^\infty \frac{1}{n^2} $$ Subtracting the RHS, we have: $$ -1 \leq \int_1^\infty \frac{1}{x^2} - \sum_{n = 1}^\infty \frac{1}{n^2}$$ Multiplying by $-1$, we have:

$$ 1 \geq \sum_{n = 1}^\infty \frac{1}{n^2} -\int_1^\infty \frac{1}{x^2} $$ as we wanted.

Answer 3

Hint: Subract $1$ from both sides to see inequality is the same as

$$\sum_{n=2}^{\infty} \frac{1}{n^2} \le \int_1^\infty\frac{dx}{x^2}.$$

Now do your rectangle comparisons.

Answer 4

More generally, suppose $f$ is strictly decreasing on $x\ge 1$, so any positive integer $n$ satisfies $f(n+1)\le\int_n^{n+1}f(x)dx\le f(n)$. Summing, $\sum_{n\ge 2}f(n)\le\int_1^\infty f(x)dx\le\sum_{n\ge 1}f(n)$. Equivalently, $\int_1^\infty f(x)dx\le\sum_{n\ge 1}f(n)\le f(1)+\int_1^\infty f(x)dx$. You just need the choice $f(x)=x^{-2}$. Another important corollary, called the integral test, is that for such $f$ the series $\sum_{n\ge 1}f(n)$ converges iff $\int_1^\infty f(x)dx$ does. In particular, the divergence of the harmonic series is equivalent to $\int_1^\infty\frac{dx}{x}=\ln\infty=\infty$.