# Negative Integer Index

## Negative Integer Index

Let m be negative integer say m = -n And |x| < 1

$${{\left( 1-x \right)}^{-n}}=1-\frac{n}{1!}x+\frac{n\left( n+1 \right)}{2!}{{x}^{2}}…+{{\left( -1 \right)}^{r}}\frac{n\left( n+1 \right)…\left( n+r-1 \right)}{r!}{{x}^{r}}+….$$.

Similarly,

$${{\left( 1+x \right)}^{n}}=\sum\limits_{r=0}^{\infty }{\,\,\,\,\,{{\,}^{n+r}}^{-1}}{{C}_{r}}{{x}^{r}}$$,

Approximations using Binomial Theorem (1 + x)ⁿ, |x| < 1,  n is any rational

$${{\left( 1+x \right)}^{n}}=1+nx+\frac{n\left( n-1 \right)}{2!}{{x}^{2}}+…+\frac{n\left( n-1 \right)\left( n-r+1 \right)}{r!}{{x}^{r}}+…$$.

• If x² and higher power can be neglected, then (1 + x)ⁿ = 1 + nx.
• (1 + x)⁻¹ if |x|< 1 = 1 – x + x² – x³ + … ∞
• (1 + x)⁻² = 1 – 2x + 3x² – 4 x³ + … ∞
• (1 + x)⁻¹ = 1 + x + x² + x³ + … ∞
• (1 + x)⁻² = 1 + 2x + 3x² + 4 x³ + … ∞
• $${{e}^{x}}=1+\frac{x}{1!}+\frac{{{x}^{2}}}{2!}+…\infty$$.

Where a > 0

$${{a}^{x}}=1+\frac{x}{1!}ln\,a+\frac{{{x}^{2}}}{2!}{{\left( lna \right)}^{2}}+…\infty$$.

Where a > 0

$$e=1+\frac{1}{1!}+\frac{1}{2!}+…\infty$$.

Logarithm series

Ln (1 + x) =$$x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}+…\infty$$.

Where -1< x 1

Ln (1 – x) =$$-x-\frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}+…\infty$$.

Note: 1 – ½ + ⅓ – ¼ + … ∞ = ln 2

$$\sum\limits_{n=1}^{\infty }{\frac{n}{n!}}=e$$.

$$\sum\limits_{n=1}^{\infty }{\frac{{{n}^{2}}}{n!}}=2e$$.

$$\sum\limits_{n=1}^{\infty }{\frac{{{n}^{3}}}{n!}}=5e$$.

$$\sum\limits_{n=1}^{\infty }{\frac{{{n}^{4}}}{n!}}=15e$$.