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
 A: Hint:
By Taylor,
$$\log 2=-\log\left(1-\frac12\right)=\sum_{n=1}^\infty\frac1{2^nn}.$$
Hence the difference is 

$\frac1{101\cdot2^{101}}+\frac1{102\cdot2^{102}}+\frac1{103\cdot2^{103}}\cdots\approx\frac1{101\cdot2^{101}}+\frac1{101\cdot2^{101}}+\frac1{101\cdot2^{103}}=\frac1{101\cdot2^{100}}$.


Addendum (technical):
If you want to express the exact value, consider the finite sum
$$s(x):=\sum_{k=1}^{100}\frac{x^n}n$$
We have $$s'(x)=\sum_{k=1}^{100}x^{n-1}=\frac{1-x^{100}}{1-x}.$$
From this,
$$s\left(\frac12\right)=\int_{x=0}^{1/2}\frac{1-x^{100}}{1-x}dx=\log2-\int_{x=0}^{1/2}\frac{x^{100}}{1-x}=\log2-B\left(\frac12;101,0\right)$$ where $B$ denotes the incomplete Beta integral.
A: $$\log2-\sum^{100}_{n=1}\frac{1}{2^nn} = \sum^{\infty}_{n=101}\frac{1}{2^nn}$$
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 $7.7 \times 10^{-33}$
