The sum 1 + (1+3)/2! + (1+3+5)/3! + (1+3+5+7)/4! + ... upto ∞ terms, is equal to

Q: Why is the first term of the first sum zero?

In the series $\sum \frac{1}{(r-2)!}$, when $r=1$, we get $\frac{1}{(-1)!}$. By convention, the factorial of a negative integer is undefined, but in the context of these series, we treat terms where the denominator factorial is negative as zero.

Q: Is there a shortcut to identify the series?

Yes! Whenever you see factorials in the denominator, always think of the Taylor series for $e^x$. The goal is to manipulate the numerator to match the denominator's factorial structure.

Q: Why did we split the term into two?

We split it because we couldn't directly identify the series for $e$ with the $r$ in the numerator. By splitting it, we transformed a complex term into two recognizable series.

Question

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

Options:

(A) $6e$

(B) $4e$

(C) $3e$

(D) $2e$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, traveler of the mathematical realms! Today, we are going to unravel a series that, at first glance, might seem like a chaotic jumble of factorials and odd numbers.

Let us look at our series: 
$$S = 1 + \frac{1+3}{2!} + \frac{1+3+5}{3!} + \dots$$

The first thing to notice is the structure of the numerators. We have $1$, then $1+3$, then $1+3+5$.

These are the sums of consecutive odd numbers. If you recall your algebra, the sum of the first $r$ odd numbers is simply $r^2$.

This is our first breakthrough. Now, our series looks much cleaner: 
$$S = \sum_{r=1}^{\infty} \frac{r^2}{r!}$$

### The Art of Simplification

Now, we face the challenge of the factorial in the denominator. We have the general term $T_r = \frac{r^2}{r!}$.

To make this manageable, we use the property of factorials: $r! = r 	imes (r-1)!$. By substituting this into our general term, we get:
$$T_r = \frac{r^2}{r 	imes (r-1)!}$$

We can cancel one $r$ from the numerator and denominator, leaving us with:
$$T_r = \frac{r}{(r-1)!}$$

This is much better, but we still have that pesky $r$ in the numerator. How do we get rid of it?

### The Elegant Decomposition

This is where the magic happens. We want to eliminate the variable $r$ from the numerator so that we are left with something that looks like the expansion of $e$.

We can rewrite $r$ as $(r-1) + 1$. Now, our term becomes:
$$T_r = \frac{(r-1) + 1}{(r-1)!}$$

By splitting this fraction, we get:
$$T_r = \frac{r-1}{(r-1)!} + \frac{1}{(r-1)!}$$

Simplifying the first part, we get:
$$T_r = \frac{1}{(r-2)!} + \frac{1}{(r-1)!}$$

Suddenly, the complexity vanishes, and we are left with two simple, recognizable terms.

### The Convergence to $e$

We are now ready to sum the series. Recall the Taylor series for the mathematical constant $e$:
$$e = \sum_{k=0}^{\infty} \frac{1}{k!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \dots$$

Our total sum is:
$$S = \sum_{r=1}^{\infty} \frac{1}{(r-2)!} + \sum_{r=1}^{\infty} \frac{1}{(r-1)!}$$

Let us evaluate these one by one. The first sum, $\sum_{r=1}^{\infty} \frac{1}{(r-2)!}$, gives us $\frac{1}{(-1)!} + \frac{1}{0!} + \frac{1}{1!} + \dots$.

Since $\frac{1}{(-1)!}$ is defined as zero, this sum is simply $e$. The second sum, $\sum_{r=1}^{\infty} \frac{1}{(r-1)!}$, gives us $\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \dots$, which is also $e$.

Adding them together, we get:
$$S = e + e = 2e$$

And there it is! A seemingly complex problem reduced to the elegant constant $e$. Keep practicing these decompositions; they are the key to mastering infinite series! The final answer is **$2e$**.

Animated Solution for Mathematics - Sequence and Series: The sum $1 + \frac{1 + 3}{2 !} + \frac{1 + 3 + 5}{3 !} + \frac{1 + 3 + 5 + 7}{4 !} + \dots$ upto $\infty$ terms, is equal to

Select Answer:

Visualized Solution

Analyze the Series Pattern

Pattern in Numerators

Pattern in Denominators

Define the General Term $T_{r}$

Simplify the General Term

Strategy for Factorial Series

Split the General Term

Final Form of $T_{r}$

Apply Summation

Recall the Exponential Series

Evaluate the First Sum

Evaluate the Second Sum

Final Calculation

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Art of Simplification

The Elegant Decomposition

The Convergence to $e$

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Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum 1+2!1+3​+3!1+3+5​+4!1+3+5+7​+… upto ∞ terms, is equal to