Trying to solve this proof by induction Series question

Question

Prove S_n-1 > 2(n^1/2 - 1) for all n ∈ Ν \ {1} where

S_n = 1/√1 + 1/√2 + 1/√3 + ... + 1/√n

 

I have set N = 2 for the basis step:

S_2-1 = 1/√1 > 2(√2 - 1) Which is true

 

For the assumption stage I am setting N = k:

S_k-1 > 2(√k - 1)

 

I am struggling to prove the final stage where I am setting N = k+1:

S_k > 2((√k+1) -1)

 

Any ideas where to go from here?

Answer 1

S_k = S_k-1 + 1/√k > 1/√k + 2(√k - 1)

 

All you need to prove is

 

1/√k + 2(√k - 1) > 2[√(k+1) - 1]

 

Or

 

1/√k + 2√k > 2√(k+1)

 

Square both sides to get

 

1/k + 4k + 4 > 4k + 4

 

Or

 

1/k > 0

 

QED

Answer 2

This has been a fun one.

 

We make use of one important observation:  S_k - S_k-1 = 1/√(k).

 

Proof:

For k>0, (k+1)/k > 1

√[(k+1)/k] > √(1) = 1

√[(k+1)/k] + 1 > 2

√[(k+1)/k]+ [√(k) / √(k)] > 2

√[(k+1)/k]+ [√(k) / √(k)] > 2(k+1-k).  Factor out 1/√(k) on the left:

[1/√(k)]*[√(k+1)+√(k)] > 2[√(k+1)-√(k)]*[√(k+1)+√(k)] 

(S_k - S_k-1)*[√(k+1)+√(k)] > 2[√(k+1)-√(k)]*[√(k+1)+√(k)] Divide both sides by [√(k+1)+√(k)]:

S_k - S_k-1 > 2[√(k+1)-√(k)]

S_k - S_k-1 > 2[√(k+1)-1]- 2[√(k)-1]
S_k > 2[√(k+1)-1]- 2[√(k)-1] + S_k-1

 

Since we have assumed S_k-1 > 2[√(k)-1]

S_k > 2[√(k+1)-1]- 2[√(k)-1] + S_k-1 > 2[√(k+1)-1]- 2[√(k)-1] + 2[√(k)-1]

 

And finally,

S_k > 2[√(k+1)-1]