Powers of 2 and harmonic numbers

Calculus Level 2

Let H n = 1 + 1 2 + 1 3 + ⋯ + 1 n H_n = 1+\frac12 + \frac13 + \cdots + \frac1{n} be the n th n^\text{th} harmonic number .

The sum ∑ n = 1 ∞ H n 2 n = ln ⁡ ( a ) \displaystyle \sum_{n=1}^\infty \frac{H_n}{2^n} = \ln (a) for some positive integer a . a.

What is a ? a?


The answer is 4.

Patrick Corn by Patrick Corn , 44, USA

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2 solutions

The serious converges by ratio test.
S = H 1 2 + H 2 4 + H 3 8 + … S = \dfrac{H_{1}}{2} + \dfrac{H_{2}}{4} + \dfrac{H_{3}}{8} + \ldots
S 2 = H 1 4 + H 2 8 + H 3 16 + … \dfrac{S}{2} =\dfrac{H_{1}}{4} + \dfrac{H_{2}}{8} + \dfrac{H_{3}}{16} + \ldots
Subtracting from the original series and using,
H n − H n − 1 = 1 n H_{n} - H_{n-1} = \dfrac{1}{n}
S 2 = ∑ n = 1 ∞ 1 n 2 n \dfrac{S}{2} = \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n2^{n}}
We know that,
ln ⁡ ( 1 1 − x ) = ∑ n = 1 ∞ x n n \ln\left(\dfrac{1}{1-x}\right) = \displaystyle \sum_{n=1}^{\infty} \dfrac{x^{n}}{n}
Putting x = 1 2 x = \dfrac{1}{2} ( which is in the radius of convergence ) we get,
S 2 = ln ⁡ ( 2 ) \dfrac{S}{2} =\ln(2)
S = 2 ln ⁡ ( 2 ) = ln ⁡ ( 4 ) S = 2\ln(2) = \ln(4)


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The generating function of H n H_n is f ( z ) = ∑ H n z n = − ln ⁡ ( 1 − z ) 1 − z f(z)=\sum H_n z^n=-\frac{\ln(1-z)}{1-z} . For z = 1 2 z=\frac{1}{2} this is ln ⁡ 4 \ln\boxed{4}

Otto Bretscher - 5 years, 1 month ago
J Joseph
Nov 1, 2016

∑ n = 1 ∞ H n x n = − l n ( 1 − x ) 1 − x \sum_{n=1}^{\infty} H_n x^{n} = \frac{-ln(1-x)}{1-x}

∑ n = 1 ∞ H n 2 n = ∑ n = 1 ∞ H n ( 1 2 ) n \sum_{n=1}^{\infty} \frac{H_n}{2^{n}} = \sum_{n=1}^{\infty}H_n\left ( \frac{1}{2} \right )^{n}

= − l n ( 1 − 1 2 ) 1 − 1 2 = \frac{-ln(1-\frac{1}{2})}{1-\frac{1}{2}}

= − 2 l n ( 1 2 ) = -2ln(\frac{1}{2})

= − 2 ( l n ( 1 ) − l n ( 2 ) ) = -2(ln(1) - ln(2))

= 2 l n ( 2 ) = 2ln(2)

= l n ( 4 ) = ln(4)

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