The sum of series 1/2! + 1/4! + 1/6! + ... is (A) (e² - 2)/e (B) (e-1)²/2e (C) (e² - 1)/2e (D) (e² -

Q: Wait, why do we use $x=1$ and $x=-1$? Why not some other value?

Great question! We use $x=1$ and $x=-1$ because they are the simplest values that allow us to generate the series $\frac{1}{n!}$ without any extra powers of $x$ cluttering the expression. If we used $x=2$, we would have $2^n$ in the numerator, which would change the series entirely. By choosing $1$ and $-1$, we keep the denominators as pure factorials, which is exactly what the problem demands.

Q: Why do we subtract 1 at the end? I thought the sum was just $\frac{e + e^{-1}}{2}$.

That is a very common trap! The sum $\frac{e + e^{-1}}{2}$ actually equals $1 + \frac{1}{2!} + \frac{1}{4!} + \dots$. Notice that extra $1$ at the beginning? That $1$ comes from the $\frac{x^0}{0!}$ term in the Taylor expansion. Since your target series starts at $\frac{1}{2!}$, that leading $1$ is an intruder. We must subtract it to get the correct answer.

Q: What if the series was $\frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + \dots$? How would I solve that?

You are thinking like a mathematician! If you wanted the odd terms, you would simply subtract the two equations ($e^1 - e^{-1}$) instead of adding them. The even terms would cancel out, and the odd terms would double. You would then divide by $2$ to get the sum. It is the exact same logic, just a different operation!

Question

The sum of series $\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots$ is

Options:

(A) $\frac{e^2 - 2}{e}$

(B) $\frac{(e-1)^2}{2e}$

(C) $\frac{e^2 - 1}{2e}$

(D) $\frac{e^2 - 1}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorer of the mathematical universe. Today, we are not just solving a problem; we are peeling back the layers of one of the most beautiful structures in calculus: the exponential series.

When you look at the series 
$$\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots$$
what do you see? To the untrained eye, it is just a sequence of numbers. But to you, it should look like a puzzle waiting to be solved. The denominators are all even factorials; this is not a coincidence, but a mathematical fingerprint.

### The Master Key: The Taylor Series

Every time you encounter a series involving factorials, your mind should immediately race to the Taylor expansion of the exponential function. The master formula is:

$$e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$$

This series is the "DNA" of the exponential function. It contains all powers of $x$ and all factorials, both odd and even. To isolate the even terms, we must use the power of symmetry.

### The Algebraic Surgery

Imagine we have two versions of this series: one where $x=1$ and one where $x=-1$. When we set $x=1$, we get:

$$e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots$$

When we set $x=-1$, the odd powers of $x$ become negative, while the even powers remain positive:

$$e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$$

Now, look at what happens when we add these two equations. The term $\frac{1}{1!}$ meets $-\frac{1}{1!}$ and vanishes, and the term $\frac{1}{3!}$ meets $-\frac{1}{3!}$ and vanishes. The odd terms are perfectly annihilated, leaving the even terms doubled:

$$e + e^{-1} = 2 \left( 1 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \right)$$

### The Final Adjustment

We have isolated the even terms, but we have an extra $1$ at the start of our series. Our target series is $\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots$, while our current expression includes the leading $1$.

To fix this, we subtract $1$ from both sides:

$$\frac{e + e^{-1}}{2} - 1 = \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots$$

Now, it is just a matter of algebraic cleanup. We rewrite $e^{-1}$ as $\frac{1}{e}$ and combine the terms:

$$\frac{e + \frac{1}{e}}{2} - 1 = \frac{e^2 + 1}{2e} - 1 = \frac{e^2 + 1 - 2e}{2e}$$

Recognizing that the numerator is a perfect square, we arrive at the final result:

$$\mathbf{\frac{(e-1)^2}{2e}}$$

### Conclusion

Mathematics is not about memorizing formulas; it is about recognizing patterns. You have just used the symmetry of the exponential function to filter an infinite series.

This technique of adding and subtracting series is a powerful tool in your arsenal. Keep this logic close to your heart, and the next time you see a series, you will see the hidden structure of the universe waiting to be revealed.

Animated Solution for Mathematics - Sequence and Series: The sum of series $\frac{1}{2 !} + \frac{1}{4 !} + \frac{1}{6 !} + \dots$ is

Select Answer:

Visualized Solution

Target Series

Exponential Series Expansion

Substitute $x = 1$

Substitute $x = - 1$

Add Equation 1 and Equation 2

Isolate the Even Series

Subtract 1 from Both Sides

Simplify the Expression

Final Conclusion

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Master Key

The Algebraic Surgery

The Final Adjustment

Conclusion

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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 of series 2!1​+4!1​+6!1​+… is