Who Let The Logarithms In?
2 3 ln 4 + 3 3 ln 6 + 4 3 ln 8 + 5 3 ln 1 0 + ⋯
You may use the approximations:
ζ
(
3
)
≈
1
.
2
0
2
0
5
6
9
and
ζ
′
(
3
)
≈
−
0
.
1
9
8
1
2
6
.
Notation : ζ ( ⋅ ) denotes the Riemann zeta function .
The answer is 0.3381.
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1 solution
S 2 equals − ζ ′ ( 3 ) . You missed the minus sign.
I found it extremely hard to use zeta'(3) in wolfram alpha, so i have written some approximation by editing this problem @Harsh Shrivastava
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If you just type
L o g [ 2 ] ( Z e t a [ 3 ] − 1 ) − Z e t a ′ [ 3 ]
into WA you will have problems, since it tries to evaluate it exactly. On the other hand, if you make sure to let WA know that you want a numerical answer, by (for example) typing (spot the decimal point)
L o g [ 2 . ] ( Z e t a [ 3 ] − 1 ) − Z e t a ′ [ 3 ]
then WA has no problems with this at all.
For that matter, WA will sum this series exactly, and give the numerical answer too, but that takes the fun out of the question!
Upside down a is what now?
The sum we want is: S = n = 2 ∑ ∞ n 3 ln ( 2 n )
Now on modifying the summation accordingly, we get: S = ( n = 1 ∑ ∞ n 3 ln ( 2 ) − ln ( 2 ) ) + n = 1 ∑ ∞ n 3 ln ( n )
Now, we take S = S 1 + S 2
Clearly, by definition of Riemann Zeta Function S 1 = ln ( 2 ) [ ζ ( 3 ) − 1 ]
Now, we define: A ( j ) = n = 1 ∑ ∞ n 3 n j = ζ ( 3 − j )
Therefore, we get S 2 = A ′ ( 0 ) = − ζ ′ ( 3 )
Therefore, S = ln ( 2 ) [ ζ ( 3 ) − 1 ] − ζ ′ ( 3 )
Now I had to take the help of Wolfram Alpha to get the answer to decimals.