What is the sum from n=0 to infinity of (j/(1+j))^n ?

The sum from n=0 to infinity of (j/(1+j))^n is converges

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sum from n=0 to infinity of (j/(1+j))^n

n = 0 \sum \infty (\frac{j}{1 + j})^{n}

converges

Solution steps

j = 0 \sum \infty (\frac{n}{1 + n})^{j}

n \to \infty lim n \frac{( \frac{j}{1 + j} ) ^{n}}{( \frac{j}{1 + j} ) ^{n + 1}} - 1 = \infty

L > 1,

by the Raabe's test

= converges

n = 1 \sum \infty \frac{e ^{\frac{4}{n}}}{n}

n = 1 \sum \infty \frac{e ^{\frac{2}{n}}}{n}

n = 1 \sum \infty \frac{1}{n ^{9}}

n = 1 \sum \infty \frac{9}{n} - \frac{9}{n + 1}

n = 1 \sum \infty \frac{6}{3 n ^{2} + 2}

What is the sum from n=0 to infinity of (j/(1+j))^n ?
The sum from n=0 to infinity of (j/(1+j))^n is converges