I am assuming this uses some kind of geometric or taylor series expansion but I am not able to see how to get it :$$\sum_k k (1-e^{-\lambda c})^k e^{-\lambda c}$$
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3Hint: can you sum $\sum\limits_kkt^k$? – Did Sep 18 '13 at 18:21
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I am not sure if I can, how would I go about doing that is the question I am asking (coming from a non mathematical background)? – knk Sep 18 '13 at 18:25
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This post should give you ideas. – David Mitra Sep 18 '13 at 18:26
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If we denote $u=1-e^{-\lambda c},$ then $$ \sum_k k (1-e^{-\lambda c})^k e^{-\lambda c}=(1-u)u\sum_k {k u^{k-1} }=(1-u)u\left(\sum_k { u^{k} }\right)'. $$
M. Strochyk
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HINT:
$$\sum_\sum_{a\le k\le b} k (1-e^{-\lambda c})^k e^{-\lambda c} =e^{-\lambda c}\sum_{a\le k\le b} k (1-e^{-\lambda c})^k $$
Let $$\sum_{a\le r\le b}A x^r=Ax^a\frac{x^{b-a+1}-1}{x-1}=A\frac{x^{b+1}-x^a}{x-1}$$
Differentiate wrt $x$ and multiply both sides by $x$
lab bhattacharjee
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