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This question already has an answer here:

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.

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marked as duplicate by Zain Patel, Jack D'Aurizio sequences-and-series Apr 23 '17 at 16:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Nooo, $\log(1+x)\leq x$ for any $x\geq 0$, hence the given series is convergent by comparison with a geometric series. $\endgroup$ – Jack D'Aurizio Apr 23 '17 at 16:36
  • $\begingroup$ your sum converges by the ration test $\endgroup$ – Dr. Sonnhard Graubner Apr 23 '17 at 16:38
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$lim_{x\rightarrow 0}{{ln(1+x)}\over x}$ 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.

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Hint : Take $\displaystyle v_n=\frac{1}{2^n}$ , and use limit comparison test by taking $\displaystyle u_n=\ln \left(1+\frac{1}{2^n}\right)$

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