Proving an inequality, Real Analysis

[moo5003]

Homework Equations

Prove the following for n > 1, n is a natural number.

Sum (from i=1 to n) of:
1/sqrt(i)

is > then

sqrt(n)

The Attempt at a Solution

To be honest I have spent 2+ hours on this with little results. I have tried induction on n, I tried showing one increases more every step then the other ie:

A_n = Sum (from i=1 to n) of:
1/sqrt(i)

B_n = sqrt(n)

That A_(n+1) / A_n > B_(n+1) / B_n

I have tried squaring both sides and simplifying the expression. I just cannot seem to make any progress on this. Any help would be appreciated.

 

[Office_Shredder]

Product or sum? If it's the product of 1/sqrt(i) then each time n increases, the product decreases (and is always less than 1)

 

[moo5003]

I'm sorry it should be a sum, I'll edit the original post.

 

[moo5003]

Anyone have a hint to even start he problem? I'm a bit stumped.

 

[Billy Bob]

Can you think of the sum as a Riemann sum that is greater than an easy integral?

 

[moo5003]

If anyone cares I ended up doing this by induction.

n=2 trivial to check

A_n = Sum (from i=1 to n): 1/sqrt(i)

A_(n+1) = A_n + 1/sqrt(n+1)

Must show: A_(n+1) > sqrt(n+1) = sqrt(n) + sqrt(n+1) - sqrt(n)

by our induction hypothesis this is if and only if:

1/sqrt(n+1) > sqrt(n+1) - sqrt(n)

iff

1 > n + 1 - sqrt(n^2 + n)

Note: sqrt(n^2 + n) > n thus our above inequality holds proving

A_(n+1) > sqrt(n+1)

 

[PCurious]

If you are familiar with the AM-HM inequality, then you could do this:

\frac{\sum^{n}_{i=1} \sqrt{i}}{n} \ge \frac{n}{\sum^{n}_{i=1} \frac{1}{\sqrt{i}}}

So

\sum^{n}_{i = 1} \frac{1}{\sqrt{i}} \ge \frac{n^2}{\sum^{n}_{i=1} \sqrt{i}} \ge \frac{n^2}{\sum^{n}_{i=1} \sqrt{n}} = \frac{n^2}{n \sqrt{n}} = \sqrt{n}

Equality occurs iff each of the square-roots in the original sum is equal, which is only the case for n = 1. In all other cases, the inequality is strict.