Infinite Series Factorial Shortcut

 

❓ Question

Evaluate:

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

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Infinite series
👉 Exponential expansion
👉 Factorial manipulation.

Main idea:

Simplify:

$$\frac{k(k+1)}{k!}$$

and connect with:

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

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🔷 Step 1 — Simplify General Term 💯

Given:

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

Now:

$$k!=k(k-1)!$$

Thus:

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

Hence:

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

Let:

$$n=k-1$$

Then:

$$k=n+1$$

So:

$$S= \sum_{n=0}^{\infty} (-1)^n \frac{n+2}{n!}$$

Split series:

$$S= \sum_{n=0}^{\infty} \frac{(-1)^n n}{n!} + 2\sum_{n=0}^{\infty} \frac{(-1)^n}{n!}$$

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🔷 Step 2 — Evaluate First Series

Since:

$$\frac{n}{n!}=\frac1{(n-1)!}$$

we get:

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

Put:

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

Using:

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

Thus:

$$= -\frac1e$$

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🔷 Step 3 — Evaluate Second Series

$$2\sum_{n=0}^{\infty}\frac{(-1)^n}{n!} = \frac2e$$

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🔷 Step 4 — Final Value

$$S = -\frac1e+\frac2e$$
$$= \boxed{ \frac1e }$$

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Direct expansion karte rehna

❌ $n/n!$ simplification miss kar dena

Remember:

$$\boxed{ \frac{n}{n!}=\frac1{(n-1)!} }$$

and always connect factorial series with:

$$e^x$$

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✅ Final Answer

$$\boxed{ \frac1e }$$

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