Find the values of x for which the series 1/x + 1/x^2 + 1/x^3+… converges and express the sum as a function of x | infinite series problems and solutions

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Find the values of x for which the series 1/x + 1/x^2  + 1/x^3+…  converges and express the sum as a function of x

\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^4}+...

Related  prove that pi/8 =1/1.3 + 1/5.7 + 1/9.11 … Or show that pi/8 =1/1.3 + 1/5.7 + 1/9.11 π/8= 1/1.3 + 1/5.7 + 1/9.11 pi series infinite series for pi | The sum of the series pi/8 =1/1.3 + 1/5.7 + 1/9.11 to ∞ is

Solution

The geometric sum formula for infinite terms is given as:

If  common ration |r| < 1, S_\infty=\frac{a}{1-r}

If |r| > 1, the series does not converge and it has no sum.

Where

a is the first term of series

r is the common ratio of series

n is the number of terms of series

This is a geometric series with ratio 1/x. It converges for |1/x|<1  that is , for |x|>1 The sum is

\frac{\frac{1}{x}}{1-\frac{1}{x}}=\frac{1}{x-1}


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