Proving summation series inequality

karan000
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Question
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Attempt
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I've attempted to use Riemann sums and use the integral to prove the inequality, not sure if this was the right approach to start with as I am now stuck and don't see what to do next.

For part (b), I know that if (2√n -2) → ∞ as n → ∞, then Sn → ∞ for n → ∞ hence the summation series is divergent.
 
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##S_n > 2 \sqrt{n}-2## and ##S_n > \sqrt{n}## are not the same.
I think you are supposed to use induction in (a). The integral approach works, but it needs more mathematics.

For part (b), I know that if (2√n -2) → ∞ as n → ∞, then Sn → ∞ for n → ∞ hence the summation series is divergent.
Good, as (2√n -2) → ∞ for n → ∞ is true.
 
does it help to shift the series from k=1 to k=2 which differ only by 1? if you can prove the inequality holds for Sn>2sqrt(n)>sqrt(n) since n>0

maybe there are cases where you can shift the sum by a real number so that the first term of the sum is equal to k in the integral 2sqrt(n)-k ? Euler proved the series convergence for k=1 to n=infinity, 1/n^2=pi^2/6
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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