<s>[INST] sum of series $\log2-\sum^{100}_{n=1}\frac{1}{2^nn}$ The difference of

$$\log2-\sum^{100}_{n=1}\frac{1}{2^nn}$$
$1). $less than Zero
$2). $Greater than  $1$
$3). $less than $\frac{1}{2^{100}101}$
$4). $greater than $\frac{1}{2^{100}101}$

Solution I tried:  we know that $$\log2 = 1-\displaystyle \frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....$$ hence the given series become $$1-\displaystyle \frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.... \;\;\;-  \left (  \frac{1}{2}+\frac{1}{8}+\frac{1}{24}+.. \right)$$
I have no idea how to proceed further; please provide a hint.
Thank you
 [/INST] $$\log2-\sum^{100}_{n=1}\frac{1}{2^nn} = \sum^{\infty}_{n=101}\frac{1}{2^nn}$$
You can put bounds on this:
$$\frac{1}{2^{101} 101} \lt \sum^{\infty}_{n=101}\frac{1}{2^nn} \lt \sum^{\infty}_{n=101}\frac{1}{2^n 101}= \frac{1}{2^{100} 101}$$ 
so between about $3.9 \times 10^{-33}$ and about $7.9 \times 10^{-33}$; in fact about $7.7 \times 10^{-33}$
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