Prove by induction on m where m is an integer, such that m ≥ 2:

$$Pm:=\sum_{n=1}^{m} \frac{1}{\sqrt{n}} < 2\sqrt{m}-1$$

I know this holds for the base case since when m=2, P2 is:

$$4<3\sqrt2$$ which is clearly true. The inductive hypothesis is suppose Pn is true for all m ≥ 2, which we will use to show Pm+1 is true for m ≥ 2, ie:

$$\sum_{n=1}^{m+1} \frac{1}{\sqrt{n}} < 2\sqrt{m+1}-1$$

The LHS of the inequality can be rewritten as: $$x=\sum_{n=1}^{m+1} \frac{1}{\sqrt{n}}= \sum_{n=1}^{m} \frac{1}{\sqrt{n}}+\frac{1}{\sqrt{m+1}}$$

Thus, the induction hypothesis implies:

$$x <\frac{1}{\sqrt{m+1}}+2\sqrt{m}-1$$

So my question, would my proof my induction be valid if I show:

$$\frac{1}{\sqrt{m+1}}+2\sqrt{m}-1< 2\sqrt{m+1}-1$$?

Working out the algebra, I found this last inequality is equivalent to the statement, "0< m+1" which is clearly true if m is a positive integer. So does this pretty much complete the proof?

  • $\begingroup$ Taking on trust the algebra you have left out, that should be fine. $\endgroup$
    – David
    Commented Jan 28, 2016 at 6:08

2 Answers 2


Cancelling out $1$ on both sides,and then multiply $\sqrt{m + 1}$ on both sides shows that the inequality to be proved is equivalent to $$1 + 2\sqrt{m(m + 1)} < 2(m + 1).$$ Further simplification gives that $$2\sqrt{m(m + 1)} < 2m + 1.$$ Square both sides: $$4m^2 + 4m < 4m^2 + 4m + 1.$$ That is, $0 < 1$, done.


$P$ a partition of $[1,m]$ $P=\{1,2,...m\}$ and $f(x)=\frac{1}{\sqrt{x}}$

$$Pm=I(f,P)<\int_{1}^{m} \frac {dx}{\sqrt{x}}=2\sqrt{m}-2< 2\sqrt{m}-1$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .