Can you please explain why ∑k=1∞k2k=2

Question

Can you please explain why  ∑k=1∞k2k=2

mauricio0815sh · Accepted Answer

Since infinite series with nonnegative terms can be rearranged arbitrarily,  ∑i=1∞i2i=∑i=1∞∑j=1i12i=∑j=1∞∑i=j∞12i=∑j=1∞12j−1=2  More graphically,  12+24+38+416+…  =12+14+18+116+…(=1)  +14+18+116+…(=12)  +18+116+…(=14)  +116+…(=18)

Philip Williams · Accepted Answer

Such a sequence is called Arithemtico-Geomteric Progression.Sn=∑i=1ni2iSn2=∑i=1ni2i+1=∑i=2n+1i−12iSubtractingSn2=∑i=1n12i−n2n+1as n→∞ It&#039;s easily seen that S∞=2How to evaluate that limit∑i=1nxi−1=1−xn1−xIf |x|&amp;lt;1 as n→∞∑i=1nxi−1=1−xn1−x=11−xSecond one is directly from Taylor series.Although there exist simpler proof , I have a rigorous proof of the second part of the limit0&amp;lt;log2x=log2e∫1x1tdtWhen x&amp;gt;11t&amp;lt;1t is validlog2e∫1x1tdt&amp;lt;log2e∫1x1tdt=2log2e(x−1)&amp;lt;2log2e⋅x0&amp;lt;log2x&amp;lt;2log2ex⇒0&amp;lt;log2xx&amp;lt;2log2ex (1)As x→∞ using (1) and squeeze principle. We getlimx→∞log2xx=0 (2)By continuity of 2t making the sutbtituion x=2t and as x→∞ then. t→∞. Now (2) is changed to

alenahelenash · Accepted Answer

Start with the geometric series s(x)=∑k=0∞xk=11−x Then xddxs(x)=∑k=1∞kxk=x(1−x)2 Your case has x=1/2

Can you please explain why \sum_{k=1}^\infty\frac{k}{2^k}=2

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