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Use summation given below to find a function that it represents
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\[\sum _{n=1}^{\infty } \text{nx}^n\]
\[x/(x-1)^2\]
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give me a second
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anything?
yeah, sorry, busy busy, here it goes
\[\sum_{1}^{∞}n x ^{n} = x * \sum_{1}^{∞}n x ^{n-1} = x * d/dx[\sum_{1}^{∞} x ^{n}] = x * d/dx[1/(1-x)] = x/[(1-x)^{2}]\] Excuse me for the poor formating but the website doesn't offer much of choice. The star (*) stands for multiplication. Thanks.
Hey there's one catch: correct this: \[x * d/dx[\sum_{0}^{n} x ^{n}]\] Because when we differentiate, n->0 not n->1 Thanks
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