Prove by mathematical induction: $\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1$

Question

Could anybody help me by checking this solution and maybe giving me a cleaner one.

Prove by mathematical induction: $$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1; n\geq2$$.

So after I check special cases for $n=2,3$, I have to prove that given inequality holds for $n+1$ case by using the given $n$ case. Ok, so this is what I've got by now:

$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{(n+1)^2}\overset{?}{>}1$$

$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1$$

$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1+\frac{1}{n}$$

From the $n$ case we know that:

$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}>1+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}$$

So we basically have to prove that:

$$1+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1+\frac{1}{n}$$

$$\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}\frac{1}{n}$$

Since

$$n^2+1<n^2+2<\dots<n^2+2n+1<2n^2+n$$ for $n\geq2$, then also:

$$\frac{1}{n^2+1}>\frac{1}{2n^2+n}$$ $$\frac{1}{n^2+2}>\frac{1}{2n^2+n}$$ . . $$\frac{1}{n^2+2n+1}>\frac{1}{2n^2+n}$$

so then we have:

$$\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}>\frac{1}{2n^2+n}+\dots+\frac{1}{2n^2+n}=(2n+1)\frac{1}{2n^2+n}=\frac{1}{n}$$

Answer 1

Your argument is fine and quite clearly presented. You can shorten the presentation considerably, though, by doing something like this:

For $n\ge 2$ let $a_n=\sum_{k=n}^{n^2}\frac1k$. Since $a_2=\frac12+\frac13+\frac14>1$, it suffices to show that $a_{n+1}\ge a_n$ for $n\ge 2$. Since $n^2+2n+1<2n^2+n$ for $n\ge 2$, we have

$$a_{n+1}-a_n=\sum_{k=1}^{2n+1}\frac1{n^2+k}-\frac1n\ge\sum_{k=1}^{2n+1}\frac1{2n^2+n}-\frac1n=\frac{2n+1}{2n^2+n}-\frac1n=0\;,$$

so $a_{n+1}\ge a_n$, and the result follows by induction.

Answer 2

Your reasoning is a little overbearing on the sums, so here's a simplification although it looks like it's basically the same thing. Define $F(n)=\sum_{k=n}^{n^2}\frac{1}{k}$. Then

$$F(n+1)=F(n)-\frac{1}{n}+\frac{1}{n^2+1}+\cdots +\frac{1}{n^2+2n+1}.$$

By the induction hypothesis, $F(n)>1$. So it suffices to prove:

$$\frac{1}{n^2+1}+\cdots +\frac{1}{n^2+2n+1}-\frac{1}{n}>0,$$

which is equivalent to:

$$\frac{1}{n^2+1}+\cdots +\frac{1}{n^2+2n+1}>\frac{1}{n}.$$

There are $2n+1$ terms on the left side, so

$$\frac{1}{n^2+1}+\cdots +\frac{1}{n^2+2n+1}\geq \frac{2n+1}{n^2+2n+1}=\frac{2n+1}{(n+1)^2}.$$

Now just show $(2n+1)/(n+1)^2 > 1/n$ by cross multiplying.

Answer 3

Here is an alternate solution. We want to prove $$ \frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1 $$ Or, equivalently, that $$ \frac{\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}}{n^2 - n + 1}>\frac{1}{n^2 - n + 1} $$ where the left-hand side is the arithmetic mean of all the fractions. We know that the arithmetic mean is greater than the harmonic mean (equality only when all terms are equal, which isn't the case since $n \geq 2$): $$ \frac{\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}}{n^2 - n + 1} > \frac{n^2 - n + 1}{n + (n+1) + \cdots + n^2}\\ = \frac{n^2 - n + 1}{\frac{n + n^2}{2}(n^2 - n + 1)} = \frac{2}{n^2 + n} $$ and the proof is completed by noting that we have $$ n^2 - 3n + 2 \geq 0\\ 2n^2 - 2n + 2 \geq n^2 + n\\ 2(n^2 - n + 1) \geq n^2 + n\\ \frac{2}{n^2 + n} \geq \frac1{n^2 - n + 1} $$

Answer 4

If you accept a proof without induction, here is one: \begin{align*} \frac 1n+\Bigl(\frac 1{n+1}+\dots+ \frac1{n^2}\Bigr)>\frac 1n +(n^2-n)\cdot\frac 1{n^2}=\frac 1n+1-\frac1n=1. \end{align*} This computation is valid if $n^2>n$, i.e. if n>1.

Answer 5

In case you want a different approach, here is one:

The red rectangles represent the sum $S_n=\frac 1n+...+\frac 1{n^2}$. Above it is for $n=2$, but the principle is general. From the figure it follows immediately that $$ \int_n^{n^2}\frac 1x\ dx=\ln(n)<\frac 1n+\frac1{n+1}+...+\frac1{n^2} $$ and for $n>e$ this implies $S_n>1$.