Question

e e (7) - log base e (49)/(22) + log base e (343)/(32)....upto infini

Answer 1

$\frac{\log_e(7)}{e}$

Answer 2

The given series is: $S = \text{e e (7)} - \frac{\log_e(49)}{2!} + \frac{\log_e(343)}{3!} - \dots \text{upto infinity}$

Let's analyze the terms of the series. The numbers inside the logarithm are powers of 7: $7^1=7$, $7^2=49$, $7^3=343$. The denominators are factorials: $2!, 3!$. The signs are alternating, starting with positive, then negative, then positive.

Let's assume the first term "e e (7)" is a typo and it should be $\log_e(7)$. This makes the series follow a consistent pattern. If the series starts from $n=1$, the general term can be written as $T_n = (-1)^{n+1} \frac{\log_e(7^n)}{n!}$.

Let's verify the terms with this assumption: For $n=1$: $T_1 = (-1)^{1+1} \frac{\log_e(7^1)}{1!} = \frac{\log_e(7)}{1!} = \log_e(7)$. This matches our assumption for the first term. For $n=2$: $T_2 = (-1)^{2+1} \frac{\log_e(7^2)}{2!} = -\frac{\log_e(49)}{2!}$. This matches the second term given in the question. For $n=3$: $T_3 = (-1)^{3+1} \frac{\log_e(7^3)}{3!} = +\frac{\log_e(343)}{3!}$. This matches the third term given in the question.

So, the series can be written as: $S = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{\log_e(7^n)}{n!}$

Now, we use the logarithm property $\log_e(a^b) = b \log_e(a)$: $S = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{n \log_e(7)}{n!}$

Factor out $\log_e(7)$ from the summation: $S = \log_e(7) \sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n!}$

Simplify the term $\frac{n}{n!}$: $\frac{n}{n!} = \frac{n}{n \times (n-1)!} = \frac{1}{(n-1)!}$

Substitute this back into the series: $S = \log_e(7) \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{(n-1)!}$

Let's write out the terms of the summation: For $n=1$: $(-1)^{1+1} \frac{1}{(1-1)!} = (-1)^2 \frac{1}{0!} = 1 \times \frac{1}{1} = 1$ For $n=2$: $(-1)^{2+1} \frac{1}{(2-1)!} = (-1)^3 \frac{1}{1!} = -1 \times \frac{1}{1} = -1$ For $n=3$: $(-1)^{3+1} \frac{1}{(3-1)!} = (-1)^4 \frac{1}{2!} = 1 \times \frac{1}{2} = \frac{1}{2!}$ For $n=4$: $(-1)^{4+1} \frac{1}{(4-1)!} = (-1)^5 \frac{1}{3!} = -1 \times \frac{1}{6} = -\frac{1}{3!}$

So the summation part is: $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{(n-1)!} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$

We know the Maclaurin series expansion for $e^x$: $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$ For $x = -1$: $e^{-1} = 1 + (-1) + \frac{(-1)^2}{2!} + \frac{(-1)^3}{3!} + \frac{(-1)^4}{4!} + \dots$ $e^{-1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$

Comparing this with the summation part of our series, we see that: $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{(n-1)!} = e^{-1}$

Substitute this back into the expression for $S$: $S = \log_e(7) \cdot e^{-1}$ $S = \frac{\log_e(7)}{e}$

The final answer is $\boxed{\frac{\log_e(7)}{e}}$.