# Comparison test of series with ln function [duplicate]

How to test convergence of series with comparison test in the next examples:

$$\sum_{n=1}^\infty \ln\left(1+\frac{1}{2^n} \right)$$

I know that: (incorrect)

$$\ln\left(1+\frac{1}{2^n}\right) < 1 + \frac{1}{2^n}$$

Correct:

$$\ln\left(1+\frac{1}{2^n}\right) \leq \frac{1}{2^n}$$

Therefore (incorrect) $$\sum_{n=1}^\infty 1+\frac{1}{2^n}\$$

Therefore:

$$\sum_{n=1}^\infty \frac{1}{2^n}\$$

Is divergent, but that doesn't tell me anything because it is larger than the starting series. So how to solve this using comparison test?

Is convergent which implies that starting series is convergent as of rule:

Suppose that $$0 \leq a_n \leq b_n$$ for sufficiently large n, then:

If $\sum_{n=1}^{\infty} a_n$ diverges, then $\sum_{n=1}^{\infty} b_n$ diverges.

If $\sum_{n=1}^{\infty} b_n$ converges, then $\sum_{n=1}^{\infty} a_n$ converges.

## marked as duplicate by Zain Patel, Jack D'Aurizio sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 23 '17 at 16:37

• Nooo, $\log(1+x)\leq x$ for any $x\geq 0$, hence the given series is convergent by comparison with a geometric series. – Jack D'Aurizio Apr 23 '17 at 16:36
$lim_{x\rightarrow 0}{{ln(1+x)}\over x}$ is the derivative of $ln(1+x)$ at $0$ and is $1$ this implies that $lim_{n\rightarrow +\infty}{{ln(1+1/2^n)}\over {1/2^n}}=1$ so the serie converges since the serie $1/2^n$ converges.
Hint : Take $\displaystyle v_n=\frac{1}{2^n}$ , and use limit comparison test by taking $\displaystyle u_n=\ln \left(1+\frac{1}{2^n}\right)$