Infinite Natural Logarithmic Factorial Summation

Calculus Level 4

$\large \dfrac{1+3}{1!}\ln3 + \dfrac{1+3^2}{2!}(\ln3)^2 + \dfrac{1+3^3}{3!}(\ln3)^3 + \ldots$

What is the value of the above expression?

The answer is 28.

1 solution

Sudhanshu Mishra
Jun 19, 2015

$S=S_1+S_2$

$1 + S_1=1+\frac{ln3}{1!}+\frac{(ln3)^{2}}{2!}+\frac{(ln3)^{3}}{3!}+........$

$1 + S_2=1+\frac{3ln3}{1!}+\frac{(3ln3)^{2}}{2!}+\frac{(3ln3)^{3}}{3!}+.........$

$1+S_1=e^{ln3} = 3$

$S_1=2$

$1+S_2=e^{3ln3}=27$

$S_2=26$

$S=S_1+S_2$

$S=28$

Hello, I too got the same answer as you, however, both of our answers are incorrect. I too made the same mistake as you as well. You write that $e^{3ln3} = 9.$ However, $e^{3ln3} = e^{ln(3^3)} = e^{ln(27)} = 27$ It was a simple arithmetic error. In the end, the sum of the series is 28, and so the sum of the digits is 10. Therefore, the answer is correct as it is.

Varun Gudibanda - 5 years, 11 months ago

Oh yes.You are right . I completely missed it.

Sudhanshu Mishra - 5 years, 11 months ago

kept entering 28 :(

Abhinav Raichur - 5 years, 11 months ago

Infinite Natural Logarithmic Factorial Summation

The answer is 28.

1 solution

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