The value of Σ (k=1 to ∞) (-1)^(k+1) k(k+1)/k! is (1) 2/e (2) 1/e (3) e/2 (4) √e

Q: Wait, why does the first summation start from $k=2$ instead of $k=1$?

That is a fantastic observation! When we simplified the first part of the series, we ended up with a term involving $\frac{1}{(k-2)!}$. If we plug in $k=1$, we get $\frac{1}{(-1)!}$, which is undefined. However, in our original expression, the term for $k=1$ was zero. By shifting the index, we effectively remove that zero term, so starting from $k=2$ is mathematically correct and necessary.

Q: Why can't we just use the Taylor series for $e^x$ directly on the original expression?

You absolutely could, but it would be much more complicated! The original expression has a product $k(k+1)$ in the numerator. Decomposing it first makes the series look like standard exponential series, which is much easier to evaluate. It is all about choosing the path of least resistance.

Q: Why is the sign of the result $1/e$ and not $-1/e$?

Great question! Let's look at the signs. The first sum $S_1$ resulted in $-e^{-1}$, and the second sum $S_2$ resulted in $2e^{-1}$. When you add them together, $-e^{-1} + 2e^{-1} = e^{-1}$. The alternating signs in the original series are handled carefully by the index shifting, leading to this positive result.

Question

The value of $\sum_{k=1}^\infty (-1)^{k+1} \frac{k(k+1)}{k!}$ is

Options:

(A) 2/e

(B) 1/e

(C) e/2

(D) $\sqrt{e}$

Sigma Percentile · Accepted Answer

### The Beauty of Infinite Series: A Journey into $e$

Imagine you are standing on the edge of an infinite abyss, looking at a mathematical structure that seems daunting at first glance:

$$\sum_{k=1}^\infty (-1)^{k+1} \frac{k(k+1)}{k!}$$

It is an alternating series, dancing between positive and negative values, and it is governed by the factorial, which grows faster than almost anything else. But fear not! We are going to tame this beast together.

### Phase 1: The Anatomy of the Term

Our first step is to understand the general term $T_k = (-1)^{k+1} \frac{k(k+1)}{k!}$. The factorial $k!$ is a powerful tool, but it is also a barrier.

We know that $k! = k(k-1)!$. By substituting this into our denominator, we see a beautiful opportunity for cancellation:

$$T_k = (-1)^{k+1} \frac{k(k+1)}{k(k-1)!}$$

The $k$ in the numerator and the $k$ in the denominator cancel out, leaving us with:

$$T_k = (-1)^{k+1} \frac{k+1}{(k-1)!}$$

This is much cleaner, but we still have that $k+1$ in the numerator.

### Phase 2: The Algebraic Surgery

To proceed, we need to make the numerator match the factorial term in the denominator. We want to see $(k-1)$ in the numerator.

We can perform a bit of algebraic surgery: $k+1 = (k-1) + 2$. Now, our term looks like this:

$$T_k = (-1)^{k+1} \frac{(k-1) + 2}{(k-1)!}$$

By splitting this into two separate fractions, we get:

$$T_k = (-1)^{k+1} \frac{k-1}{(k-1)!} + (-1)^{k+1} \frac{2}{(k-1)!}$$

This is the breakthrough! We have transformed one complex series into two manageable ones.

### Phase 3: The Exponential Connection

Now, we recall our most powerful tool: the Maclaurin series for $e^x$, which is:

$$\sum_{n=0}^\infty \frac{x^n}{n!}$$

For $x=-1$, this becomes:

$$e^{-1} = \sum_{n=0}^\infty \frac{(-1)^n}{n!}$$

Let us evaluate our two sums. The first sum, after simplifying $\frac{k-1}{(k-1)!}$ to $\frac{1}{(k-2)!}$, becomes $\sum_{k=2}^\infty \frac{(-1)^{k+1}}{(k-2)!}$.

By letting $n = k-2$, this transforms into $-\sum_{n=0}^\infty \frac{(-1)^n}{n!}$, which is simply $-e^{-1}$.

The second sum is $2 \sum_{k=1}^\infty \frac{(-1)^{k+1}}{(k-1)!}$. By letting $m = k-1$, this becomes $2 \sum_{m=0}^\infty \frac{(-1)^m}{m!}$, which is $2e^{-1}$.

### The Final Elegance

When we combine these two results, we get:

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

It is truly elegant how the complexity of the original series collapses into such a simple, fundamental constant.

Remember, whenever you face a series with factorials, look for ways to decompose the numerator. It is the key to unlocking the beauty hidden within. The final answer is **$\frac{1}{e}$**.

Animated Solution for Mathematics - Sequence and Series: The value of $\sum_{k = 1}^{\infty} (- 1)^{k + 1} \frac{k ( k + 1 )}{k !}$ is

Select Answer:

Visualized Solution

The Given Infinite Series

Simplifying the General Term

Strategy for Decomposition

Splitting into Two Summations

Further Simplification of the First Sum

Recalling the Exponential Series

Evaluating the First Summation

Evaluating the Second Summation

Combining the Results

Final Answer and Key Takeaway

The Sigma Insight: Sum of Special Series

The Beauty of Infinite Series

Phase 1

Phase 2

Phase 3

The Final Elegance

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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 value of ∑k=1∞​(−1)k+1k!k(k+1)​ is