Answer to Question #149874 in Algebra for Dhruv bartwal

Question

Apply weierstrass' inequality to prove that
[nΣi=1 (1/√i)] ≤ 1/√n! [nΠi=2 (√i-1)] +2[nΣi=2 (1/√i)]

Answer 1

$Σ^n_{i=1} (1/√i)≤ 1/√n! [Π^n_{i=2} (√i-1)] +2[Σ^n_{i=2} (1/√i)]$

For i=2:

$1+1/\sqrt{2}=(1/\sqrt2)(\sqrt2-1)+2/\sqrt{2}$

Weierstrass' inequality:

$\Pi^n_{i=1}(1-a_i)\geq1-\Sigma^n_{i=1}a_i$

$a_i\in[0..1]$

Let $a_i=1/\sqrt{i}$

Then:

$Σ^n_{i=1}a_i≤ 1/√n! [Π^n_{i=2} (1/a_i-1)] +2Σ^n_{i=2}a_i$

$1≤ 1/√n! [Π^n_{i=2} (\frac{1-a_i}{a_i})] +Σ^n_{i=2}a_i$

$1-Σ^n_{i=2}a_i≤ \frac{1}{\sqrt{n!}} Π^n_{i=2} (\frac{1-a_i}{a_i})$

But, since

$\frac{1}{\sqrt{n!}\Pi^n_{i=2}a_i}=\frac{\Pi^n_{i=2}\sqrt{i}}{\sqrt{n!}}=1$

we get Weierstrass' inequality.