# $\sum_{n=0}^{\infty} n \left( \frac{2}{3} \right)^n = ?$ [duplicate]

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$$\sum_{n=0}^{\infty} n \left( \frac{2}{3} \right)^n = ?$$

How to find it? If it lacked n before fraction, I would use formula for the sum of geometric series.

## marked as duplicate by user61527, lab bhattacharjee calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 5 '14 at 10:04

Consider the infinite geometric progression for $|x|<1$, $$\sum_{n=0}^\infty x^n=\frac1{1-x}.\tag1$$ Differentiating $(1)$ with respect to $x$ yields $$\sum_{n=0}^\infty nx^{n-1}=\frac1{(1-x)^2}.\tag2$$ Multiplying $(2)$ by $x$ yields $$\sum_{n=0}^\infty nx^{n}=\frac{x}{(1-x)^2}.\tag3$$ Setting $x=\dfrac23$ to $(3)$ yields $$\sum_{n=0}^\infty n \left(\frac23\right)^{n}=\frac{\frac23}{\left(1-\frac23\right)^2}=\large\color{blue}{6}.$$
• (1) is true for $\;|x|<1\;$ – Timbuc Jul 5 '14 at 10:03