I need to prove by induction the following. The left hand side of the inequality is basically a harmonic serie where $H(1065^k)$. I've been able to complete this problem with $H(2^k)\ge 1 + \frac k2$, but can't seen to complete this problem...

$$\sum_{k=1}^{1065^k} \frac 1k \ge 1 + \frac {1064}{1065} k$$

  • $\begingroup$ Having $k$ as a dummy variable and also as a part of the answer is bad practice. Better to write $$\sum_{r=1}^{1065^k}{1\over r}\ge1+{1064\over1065}k$$ $\endgroup$ – Gerry Myerson Nov 14 '19 at 1:01
  • $\begingroup$ Can you show $$\sum_{1065^{k-1}+1}^{1065^k}{1\over r}\ge{1064\over1065}$$ $\endgroup$ – Gerry Myerson Nov 14 '19 at 1:03
  • $\begingroup$ I actually wrote i=1 1/i I have no idea why I wrote k on both sides haha $\endgroup$ – Ian Peng Nov 14 '19 at 1:03
  • $\begingroup$ I can show $$ \ge \frac{1}{1065}$$ but I don't know about $$ \frac {1064}{1065} $$ $\endgroup$ – Ian Peng Nov 14 '19 at 1:08
  • $\begingroup$ How many terms are in the sum, and what's the smallest of those terms? $\endgroup$ – Gerry Myerson Nov 14 '19 at 1:09

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