I have the sum $$A_n:=\sum_{k=n+1}^{2n}\frac1k $$
Now I have to find all numbers $n\in \mathbb{N},$ for which $A_n >\frac35$
How do I start the proof? I assume I have to use induction, but I have only done basic induction proofs to prove equalities.
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3Hint: what is $A_{n+1} - A_n$? How can you use this for your induction? – Gnuk Oct 20 '16 at 18:03
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By recalling that $H_n$ is defined as $\sum_{k=1}^{n}\frac{1}{k}$, we have $$ A_n = H_{2n}-H_n \tag{1}$$ hence: $$ A_{n+1}-A_n = \frac{1}{2n+2}+\frac{1}{2n+1}-\frac{1}{n+1} = \frac{1}{(2n+1)(2n+2)}\tag{2} $$ implying that the sequence $\{A_n\}_{n\geq 1}$ is increasing. By direct inspection of $A_1,A_2,A_3$, we have that $ A_n> \frac{3}{5}$ as soon as $n\geq 3$.
Jack D'Aurizio
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Thanks for the detailed answer! I understand the concept overall, but first of all: Why is $H_{2n} = 1/(2n+1)$? Shouldn't it be $1/2n$? (Same question arises here for $H_n$ for me) – AxiomaticApproach Oct 21 '16 at 13:47