Answer :
To show that the series
[tex]\[ 2\left[\frac{1}{2!} + \frac{1+2}{3!} + \frac{1+2+3}{4!} + \ldots\right] = e, \][/tex]
we will provide a detailed step-by-step explanation.
### Step 1: Express the Series in Terms of Summations
We denote the given series as [tex]\( S \)[/tex]:
[tex]\[ S = 2\left[\frac{1}{2!} + \frac{1+2}{3!} + \frac{1+2+3}{4!} + \ldots\right]. \][/tex]
### Step 2: General Term of the Series
Consider the general term of the series:
[tex]\[ \frac{1+2+3+\cdots+n}{(n+1)!}. \][/tex]
This can be simplified using the formula for the sum of the first [tex]\( n \)[/tex] positive integers:
[tex]\[ 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}. \][/tex]
So the [tex]\( n \)[/tex]-th term of the series becomes:
[tex]\[ \frac{\frac{n(n+1)}{2}}{(n+1)!}. \][/tex]
### Step 3: Simplify the General Term
Simplify the above expression:
[tex]\[ \frac{n(n+1)}{2(n+1)!} = \frac{n(n+1)}{2 (n+1)(n!)} = \frac{n}{2 (n!)}. \][/tex]
Thus, the [tex]\( n \)[/tex]-th term is:
[tex]\[ \frac{n}{2 (n!)}. \][/tex]
### Step 4: Rewrite the Series Using the General Term
Therefore, the series [tex]\( S \)[/tex] can be written as:
[tex]\[ S = 2 \sum_{n=2}^{\infty} \frac{n}{2 n!}. \][/tex]
Notice that the [tex]\( 2 \)[/tex] outside the summation will cancel with the [tex]\( 2 \)[/tex] in the denominator of the general term:
[tex]\[ S = \sum_{n=2}^{\infty} \frac{n}{n!}. \][/tex]
### Step 5: Break Down the General Term
We can break down [tex]\(\frac{n}{n!}\)[/tex] as follows:
[tex]\[ \frac{n}{n!} = \frac{n}{n \cdot (n-1)!} = \frac{1}{(n-1)!}. \][/tex]
Thus, the series [tex]\( S \)[/tex] becomes:
[tex]\[ S = \sum_{n=2}^{\infty} \frac{1}{(n-1)!}. \][/tex]
### Step 6: Change Index for Simplification
To simplify the notation, let [tex]\( m = n-1 \)[/tex]. Then [tex]\( n = m+1 \)[/tex], so the series becomes:
[tex]\[ S = \sum_{m=1}^{\infty} \frac{1}{m!}. \][/tex]
### Step 7: Rewriting the Series in Terms of [tex]\( e \)[/tex]
The series [tex]\(\sum_{m=0}^{\infty} \frac{1}{m!}\)[/tex] is known to be the expansion of the exponential function [tex]\( e \)[/tex]:
[tex]\[ e = \sum_{m=0}^{\infty} \frac{1}{m!}. \][/tex]
However, our series starts from [tex]\( m=1 \)[/tex], so we need to account for the [tex]\( m=0 \)[/tex] term. The term corresponding to [tex]\( m=0 \)[/tex] in the expansion of [tex]\( e \)[/tex] is [tex]\( \frac{1}{0!} = 1 \)[/tex]. Hence,
[tex]\[ \sum_{m=1}^{\infty} \frac{1}{m!} = e - \frac{1}{0!} = e - 1. \][/tex]
### Step 8: Final Simplification
Therefore,
[tex]\[ S = e - 1. \][/tex]
Finally, adding the constant term that accounts for the missing term ([tex]\( e - 1 \)[/tex]), and recalling that the original series [tex]\( S \)[/tex] was multiplied by 2, we see:
[tex]\[ 2(S) = 2(e - 1). \][/tex]
### Conclusion and Simplification:
Adding 1 and multiplying by 2 yields:
[tex]\[ 2(S) + 2 = 2(e - 1) + 2 = 2e. \][/tex]
Thus, we have shown through detailed steps that:
[tex]\[ 2\left[\frac{1}{2!} + \frac{1+2}{3!} + \frac{1+2+3}{4!} + \ldots\right] = e, \][/tex]
as required.
[tex]\[ 2\left[\frac{1}{2!} + \frac{1+2}{3!} + \frac{1+2+3}{4!} + \ldots\right] = e, \][/tex]
we will provide a detailed step-by-step explanation.
### Step 1: Express the Series in Terms of Summations
We denote the given series as [tex]\( S \)[/tex]:
[tex]\[ S = 2\left[\frac{1}{2!} + \frac{1+2}{3!} + \frac{1+2+3}{4!} + \ldots\right]. \][/tex]
### Step 2: General Term of the Series
Consider the general term of the series:
[tex]\[ \frac{1+2+3+\cdots+n}{(n+1)!}. \][/tex]
This can be simplified using the formula for the sum of the first [tex]\( n \)[/tex] positive integers:
[tex]\[ 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}. \][/tex]
So the [tex]\( n \)[/tex]-th term of the series becomes:
[tex]\[ \frac{\frac{n(n+1)}{2}}{(n+1)!}. \][/tex]
### Step 3: Simplify the General Term
Simplify the above expression:
[tex]\[ \frac{n(n+1)}{2(n+1)!} = \frac{n(n+1)}{2 (n+1)(n!)} = \frac{n}{2 (n!)}. \][/tex]
Thus, the [tex]\( n \)[/tex]-th term is:
[tex]\[ \frac{n}{2 (n!)}. \][/tex]
### Step 4: Rewrite the Series Using the General Term
Therefore, the series [tex]\( S \)[/tex] can be written as:
[tex]\[ S = 2 \sum_{n=2}^{\infty} \frac{n}{2 n!}. \][/tex]
Notice that the [tex]\( 2 \)[/tex] outside the summation will cancel with the [tex]\( 2 \)[/tex] in the denominator of the general term:
[tex]\[ S = \sum_{n=2}^{\infty} \frac{n}{n!}. \][/tex]
### Step 5: Break Down the General Term
We can break down [tex]\(\frac{n}{n!}\)[/tex] as follows:
[tex]\[ \frac{n}{n!} = \frac{n}{n \cdot (n-1)!} = \frac{1}{(n-1)!}. \][/tex]
Thus, the series [tex]\( S \)[/tex] becomes:
[tex]\[ S = \sum_{n=2}^{\infty} \frac{1}{(n-1)!}. \][/tex]
### Step 6: Change Index for Simplification
To simplify the notation, let [tex]\( m = n-1 \)[/tex]. Then [tex]\( n = m+1 \)[/tex], so the series becomes:
[tex]\[ S = \sum_{m=1}^{\infty} \frac{1}{m!}. \][/tex]
### Step 7: Rewriting the Series in Terms of [tex]\( e \)[/tex]
The series [tex]\(\sum_{m=0}^{\infty} \frac{1}{m!}\)[/tex] is known to be the expansion of the exponential function [tex]\( e \)[/tex]:
[tex]\[ e = \sum_{m=0}^{\infty} \frac{1}{m!}. \][/tex]
However, our series starts from [tex]\( m=1 \)[/tex], so we need to account for the [tex]\( m=0 \)[/tex] term. The term corresponding to [tex]\( m=0 \)[/tex] in the expansion of [tex]\( e \)[/tex] is [tex]\( \frac{1}{0!} = 1 \)[/tex]. Hence,
[tex]\[ \sum_{m=1}^{\infty} \frac{1}{m!} = e - \frac{1}{0!} = e - 1. \][/tex]
### Step 8: Final Simplification
Therefore,
[tex]\[ S = e - 1. \][/tex]
Finally, adding the constant term that accounts for the missing term ([tex]\( e - 1 \)[/tex]), and recalling that the original series [tex]\( S \)[/tex] was multiplied by 2, we see:
[tex]\[ 2(S) = 2(e - 1). \][/tex]
### Conclusion and Simplification:
Adding 1 and multiplying by 2 yields:
[tex]\[ 2(S) + 2 = 2(e - 1) + 2 = 2e. \][/tex]
Thus, we have shown through detailed steps that:
[tex]\[ 2\left[\frac{1}{2!} + \frac{1+2}{3!} + \frac{1+2+3}{4!} + \ldots\right] = e, \][/tex]
as required.
