# Proof by induction

## Homework Statement

Prove this by induction

n
∑(1/√n) > 2(√(n+1) -1)
i=1

## The Attempt at a Solution

I have tried for several hours, with no real progress. When i assume that it works for n=k, and then try to prove for n=k+1 i get 2(√(k+2)-1)... i don't know how to partition this expression... can someone point me in the right direction....

Ray Vickson
Homework Helper
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## Homework Statement

Prove this by induction

n
∑(1/√n) > 2(√(n+1) -1)
i=1

## The Attempt at a Solution

I have tried for several hours, with no real progress. When i assume that it works for n=k, and then try to prove for n=k+1 i get 2(√(k+2)-1)... i don't know how to partition this expression... can someone point me in the right direction....

Are you absolutely required to use induction? If not, there is a quick proof using elementary calculus (specifically, definite integration).

The task is part of the induction chapter in the textbook.

Are you sure that the left-hand side of the inequality is...

$\sum_{k=1}^{n} \frac 1{\sqrt {n}}$, rather than $\sum_{k=1}^{n} \frac 1{\sqrt {k}}$ ?

Are you sure that the left-hand side of the inequality is...

$\sum_{k=1}^{n} \frac 1{\sqrt {n}}$, rather than $\sum_{k=1}^{n} \frac 1{\sqrt {k}}$ ?

woops!, i can correct it...

m
∑(1/√n) > 2(√(n+1) -1)
n=1

Again, the problem for me lies in that i can't partition(divide in to a sum) the expression on the right...

m
∑(1/√n) > 2(√(n+1) -1)
n=1

Um, what is m?

So sorry, I'm not so good at the different placements for the different sum-limits/-startpoints. I first introduced "k" when i assumed that the inequality actually was valid, and then i set n=k+1 and tried to prove it...

This should be right;
n
∑(1/√i) > 2(√(n+1) -1)
i=1

Um, what is m?

forget m :)

I think you meant to write...

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

If this is the case, here's a way to start the inductive step:

$\sum_{n=1}^{m+1} \frac 1{\sqrt{n}} = ( \sum_{n=1}^{m} \frac 1{\sqrt{n}}) + \frac 1{\sqrt{m+1}}$

no, it is 2(√(n+1) -1) = 2√(n+1) -2

Let me know if you'd like more hints, but I'm pretty sure you can finish the problem from here ;)

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no, it is 2(√(n+1) -1) = 2√(n+1) -2

Oh... the right hand side is a product. Oops.

That's ok. The steps in the proof won't change much.

$\sum_{n=1}^{m+1} \frac 1{\sqrt{n}} = ( \sum_{n=1}^{m} \frac 1{\sqrt{n}}) + \frac 1{\sqrt{m+1}}$
I already knew this, but how am i supposed to compare this result to the right side? the expression on the right side is impossible to write as a sum of any sort...? At least : 2√(m+2)...

Starting from the Left-hand side of the inequality you wish to prove (P(m+1)),

$\sum_{n=1}^{m+1} \frac 1{\sqrt{n}} = \sum_{n=1}^{m} \frac 1{\sqrt{n}} + \frac 1{\sqrt{m+1}}$.

Recall that you assumed P(m) to be true, that is

$\sum_{n=1}^{m} \frac 1{\sqrt{n}} > 2 \cdot ((\sqrt{m+1}) -1)$,

So recognizing this, you have...

$\sum_{n=1}^{m+1} \frac 1{\sqrt{n}} = \sum_{n=1}^{m} \frac 1{\sqrt{n}} + \frac 1{\sqrt{m+1}} > 2 \cdot ((\sqrt{m+1}) -1) + \frac 1{\sqrt{m+1}}$.

If i contract the right side i get: (2m +3 -2√(m+1))/√(m+1)

Don't see how this is gonna help.. :)

dont i have to show that (2m +3 -2√(m+1))/√(m+1) can be written as 2(√(m+2) -1) to prove it...?

or maybe not since its an inequality...?

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dont i have to show that (2m +3 -2√(m+1))/√(m+1) can be written as 2(√(m+2) -1) to prove it...?

or maybe not since its an inequality...?

If you plug in some numbers you'll see...

$2 \cdot (\sqrt{m+1} - 1) + \frac 1{\sqrt{m+1}} \neq 2 \cdot (\sqrt {(m+1)+1} - 1)$.

You're next step should be to show that...

$2 \cdot (\sqrt{m+1} - 1) + \frac 1{\sqrt{m+1}} > 2 \cdot (\sqrt {(m+1)+1} - 1)$.

Ray Vickson
Homework Helper
Dearly Missed
$\sum_{n=1}^{m+1} \frac 1{\sqrt{n}} = ( \sum_{n=1}^{m} \frac 1{\sqrt{n}}) + \frac 1{\sqrt{m+1}}$
I already knew this, but how am i supposed to compare this result to the right side? the expression on the right side is impossible to write as a sum of any sort...? At least : 2√(m+2)...

Your problem is to show that
$$2 \sqrt{n+1}-2 + \frac{1}{\sqrt{n+1}} \geq 2 \sqrt{n+2} - 2,\\ \text{or}\\ \sqrt{n+2} - \sqrt{n+1} \leq \frac{1}{2 \sqrt{n+1}}$$
If I were doing it I would write
$$\sqrt{n+2} = \sqrt{n+1} \left( 1 + \frac{1}{n+1} \right)^{1/2}$$
and go on from there.

yes, i got 9/4 > 8/4! or 9>8 thanks!, now its just about proving the basecase and its finished?

Ray Vickson
Homework Helper
Dearly Missed
yes, i got 9/4 > 8/4! or 9>8 thanks!, now its just about proving the basecase and its finished?

Which message are you responding to? Please use the "Quote" button, so that the discussion can be followed accurately.

Which message are you responding to? Please use the "Quote" button, so that the discussion can be followed accurately.

both of them :) the inequality that both of you posted implies that 9>8. Since this is true, the inequality you posted must be true ?

Ray Vickson