Proof by induction of summation inequality: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{3^n}\ge 1+\frac{2n}3$ Proof by induction of summation inequality: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{3^n}\ge 1+\frac{2n}3$
We start with $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac1{3^n}\ge 1+\frac{2n}3$$ for all positive integers.
I have finished the basic step, the hypothesis is easy too, but I do not know what to do for n+1:
$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac1{3^n}+\frac1{3^{n+1}} \ge 1+\frac{2n}3 +\frac{1}{3^{n+1}}$$
Can you help me from here?

The sum on the LHS is
$$S_n=\sum_{k=1}^{3^n} \frac1k$$
The hypothesis is that $S_n\ge 1+\frac{2n}{3}$
Base case $n=0$. As $S_0=1\ge1+0=1$, the case $n=0$ is true.
Assume true for $n$, prove
$$S_n+\sum_{k=3^n+1}^{3^{n+1}}\frac1k\ge 1+\frac{2n}{3}+\frac{2}{3}$$
As
$$\sum_{k=3^n+1}^{3^{n+1}}\frac1k\ge\sum_{k=3^n+1}^{3^{n+1}}\frac1{3^{n+1}}=\frac{2\cdot3^n}{3^{n+1}}=\frac23$$
is true at $n=0$, the induction holds.
