Sum of series $\log2-\sum^{100}_{n=1}\frac{1}{2^nn}$

Question

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

Answer 1

Hint:

By Taylor,

$$\log 2=-\log\left(1-\frac12\right)=\sum_{n=1}^\infty\frac1{2^nn}.$$

Hence the difference is

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

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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.