Csdrhrt

I am supposed to investigate convergence of a series $ sum_{n=1 }^{infty} frac{1}{2^{n}-1} $ . I decided that the series converge to 0, because it clearly visible from first 5 elements. To do a proof, I used limit comparison test with $sum_{n=1 }^{infty} frac{1}{2^{n}} $ . Is my solution correct? Thanks

That's right when you change the sequence (for example compare it with $1over (1.5)^n$). You may also apply ratio test as following $$lim_{nto infty}{a_{n+1}over a_n}=lim_{ntoinfty}{2^{n}-1over 2^{n+1}-1}={1over 2}<1$$therefore the series converges though not to zero.

Yes we can use limit comparison test indeed

$$frac{frac{1}{2^{n}-1}}{frac{1}{2^{n}}}=frac{2^n}{2^{n}-1}to 1$$

As an alternative since for $nge 1$

$$frac{1}{2^{n}-1}lefrac{1}{2^{n-1}}$$

then

$$sum_{n=1 }^{infty} frac{1}{2^{n}-1}=1+sum_{n=2 }^{infty} frac{1}{2^{n}-1} le 1+sum_{n=2 }^{infty} frac{1}{2^{n-1}}=1+sum_{n=1 }^{infty} frac{1}{2^{n}}$$

Yes it is correct: the general term of your series
$$frac1{2^n-1}sim_{nto infty}frac1{2^n}$$
in the sense of *asymptotic analysis, i.e. $ ;dfrac{tfrac1{2^n-1}}{frac1{2^n}} $ tends to $1$ as $ntoinfty$, and two series with equivalent (positive) general term both converge or both diverge.

The series is absolutely convergent by comparison with a geometric series. We may notice that
$$ sum_{ngeq 1}frac{1}{2^n-1}=sum_{ngeq 1}sum_{kgeq 1}2^{-kn} = sum_{mgeq 1}frac{d(m)}{2^m}=frac{25}{16}+sum_{mgeq 7}frac{d(m)}{2^m} $$
and since for any $mgeq 7$ we have $2leq d(m) leq frac{m}{2}$ it follows that
$$ frac{51}{32}leq sum_{ngeq 1}frac{1}{2^n-1}leq frac{13}{8} $$
so
$$ sum_{ngeq 1}frac{1}{2^n-1}=frac{103}{64}+E,qquad |E|leqfrac{1}{64}. $$