How do you find the sum of the series #1+ln2+(((ln2)^2)/(2!))+...+(((ln2)^2)/(n!))+...#?

Question

Paisley Johnson · Accepted Answer

undefined To find the sum of the series $1 + \ln(2) + \frac{{(\ln(2))^2}}{{2!}} + \ldots + \frac{{(\ln(2))^n}}{{n!}} + \ldots$, you can use the Taylor series expansion of the natural logarithm function. The Taylor series expansion of $\ln(1+x)$ is $\sum_{n=1}^{\infty} \frac{{(-1)^{n+1}x^n}}{{n}}$. Substituting $x = 1$ into this series, you get $\ln(1+1) = \ln(2) = \sum_{n=1}^{\infty} \frac{{(-1)^{n+1}}}{{n}}$. This gives the sum of the series as $\ln(2)$. Therefore, the sum of the given series is $\ln(2)$.

Cameron Alexander · Answer

The sum is #2#. Let's write the McLaurin Series of #y=2^x#: #2^x=2^0+(2^0ln2)/(1!)x+(2^0(ln2)^2)/(2!)x^2+...+(2^0(ln2)^n)/(n!)x^n+...#. If we put in this series #x=1#, on the left the result is #2#, on the right the result is the series of the exercise.

Evelyn Agard · Answer

undefined To find the sum of the series $1 + \ln(2) + \frac{{(\ln(2))^2}}{{2!}} + \frac{{(\ln(2))^3}}{{3!}} + \ldots + \frac{{(\ln(2))^n}}{{n!}} + \ldots$, we can recognize it as the Maclaurin series expansion of $e^x$ evaluated at $x = \ln(2)$.

The Maclaurin series expansion of $e^x$ is given by:

$$e^x = 1 + x + \frac{{x^2}}{{2!}} + \frac{{x^3}}{{3!}} + \ldots + \frac{{x^n}}{{n!}} + \ldots$$

Substituting $x = \ln(2)$ into the Maclaurin series expansion of $e^x$, we get:

$$e^{\ln(2)} = 1 + \ln(2) + \frac{{(\ln(2))^2}}{{2!}} + \frac{{(\ln(2))^3}}{{3!}} + \ldots + \frac{{(\ln(2))^n}}{{n!}} + \ldots$$

Since $e^{\ln(2)} = 2$, the sum of the series is $2$.