How can deleting specific digits affect the bounds on the harmonic series?

[quddusaliquddus]

OK guys. Sorry to bring up an old topic. I've struggled with this. Can someone please explain to me how you get:
====================================
n
log_2(2n)>=sigma(1/n) >=log_2(n)
i=1

2^n
since 1/2 <= sigma(1/n)<=1
i=(2^n)-1
====================================

Please?

 

[NateTG]

OK guys. Sorry to bring up an old topic. I've struggled with this. Can someone please explain to me how you get:
====================================
n
log_2(2n)>=sigma(1/n) >=log_2(n)
i=1

2^n
since 1/2 <= sigma(1/n)<=1
i=(2^n)-1
====================================

Please?

Quick and dirty bounds:
\sum_{i=2^n}^{2^{n+1}} \frac{1}{2^n} \geq \sum_{i=2^n}{2^{n+1}} \frac{1}{n} \geq \sum_{i=2^n}^{2^{n+1}} \frac{1}{2^{n+1}}
The bounds both simplify to multiplication
1 \geq \sum_{i=2^n}{2^{n+1}} \frac{1}{n} \geq \frac{1}{2}
so
n+1 \geq \sum_{i=1}{2^{n+1}} \geq \frac{n}{2}
You should be able to get it from there.