Question

The sum $ 1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + ... $ upto $ \infty $ terms, is equal to

• A. \( 6e \)
• B. \( 4e \)
• C. \( 3e \)
• D. \( 2e \)

Hint:

Identify the general term of the series. In this case, the numerator is the sum of the first \( n \) odd numbers, which is \( n^2 \). Simplify the expression \( \frac{n^2}{n!} \) by writing \( n^2 = n(n-1) + n \) and splitting the sum into simpler series that can be related to the Taylor series expansion of \( e^x \) at \( x = 1 \), which is \( e = \sum_{n=0}^{\infty} \frac{1}{n!} \).

Answer 1

The Correct Option is D

To solve the series \( 1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + \ldots \) up to infinity, let’s break it down step by step.

The pattern in the series appears to be an arithmetic sequence numerator over factorial denominators. Specifically, each term can be expressed as:

• First term: \( 1 \)
• Second term: \( \frac{1 + 3}{2!} = \frac{4}{2!} \)
• Third term: \( \frac{1 + 3 + 5}{3!} = \frac{9}{3!} \)
• Fourth term: \( \frac{1 + 3 + 5 + 7}{4!} = \frac{16}{4!} \)

Observe that the sequence of the numerators \( 1, 4, 9, 16, \ldots \) follows \( n^2 \) where \( n \) is the term index: \( 1=1^2, 4=2^2, 9=3^2, 16=4^2, \ldots \).

Hence, the general term is:

\[ \frac{n^2}{n!} \]

Thus, the series becomes:

\[ \sum_{n=1}^{\infty} \frac{n^2}{n!} \]

Now, let's evaluate the sum:

We know an essential series expansion for \( e^x \), which is:

\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]

Considering \( x = 1 \), it becomes:

\[ e = \sum_{n=0}^{\infty} \frac{1^n}{n!} = \sum_{n=0}^{\infty} \frac{1}{n!} \]

To compute \(\sum_{n=1}^{\infty} \frac{n^2}{n!}\), we can derive from properties of series that involve similar factorial terms.

In fact, let's consider the function:

\[ f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x \]

Differentiate \( f(x) \) with respect to \( x \):

\[ f'(x) = \sum_{n=1}^{\infty} \frac{nx^{n-1}}{n!} = e^x \]

Multiplying by \( x \) and differentiating again will provide the relationship to resolve:

\[ x \sum_{n=1}^{\infty} \frac{n^2 x^{n-1}}{n!} = xe^x \]

Simplifying our specific requirement \( x = 1 \):

\[ \sum_{n=1}^{\infty} \frac{n^2}{n!} = e + e \quad \text{(from deriving twice and leveraging )} = 2e \]

Thus, the sum of the given infinite series is indeed: \(\boxed{2e}\)

Answer 2

The \( n^{th} \) term of the series (for \( n \ge 1 \)) is given by: \[ T_n = \frac{1 + 3 + 5 + \dots + (2n - 1)}{n!} \] The sum of the first \( n \) odd numbers is \( n^2 \). So, the numerator is \( n^2 \). \[ T_n = \frac{n^2}{n!} \] The given sum \( S \) can be written as: \[ S = 1 + \sum_{n=2}^{\infty} \frac{n^2}{n!} = \sum_{n=1}^{\infty} \frac{n^2}{n!} \] We can write \( n^2 = n(n - 1) + n \). \[ S = \sum_{n=1}^{\infty} \frac{n(n - 1) + n}{n!} = \sum_{n=1}^{\infty} \frac{n(n - 1)}{n!} + \sum_{n=1}^{\infty} \frac{n}{n!} \] For the first sum, the terms for \( n = 1 \) are zero. So, we start from \( n = 2 \): \[ \sum_{n=2}^{\infty} \frac{n(n - 1)}{n!} = \sum_{n=2}^{\infty} \frac{n(n - 1)}{n(n - 1)(n - 2)!} = \sum_{n=2}^{\infty} \frac{1}{(n - 2)!} \] Let \( m = n - 2 \). When \( n = 2 \), \( m = 0 \). When \( n \rightarrow \infty \), \( m \rightarrow \infty \). \[ \sum_{m=0}^{\infty} \frac{1}{m!} = e \] For the second sum: \[ \sum_{n=1}^{\infty} \frac{n}{n!} = \sum_{n=1}^{\infty} \frac{n}{n(n - 1)!} = \sum_{n=1}^{\infty} \frac{1}{(n - 1)!} \] Let \( k = n - 1 \). When \( n = 1 \), \( k = 0 \). When \( n \rightarrow \infty \), \( k \rightarrow \infty \). \[ \sum_{k=0}^{\infty} \frac{1}{k!} = e \] Therefore, the sum \( S \) is: \[ S = e + e = 2e \]