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Category: Relation and Functions

1-4-16-2-64-6-4-n-n-

Question Number 160136 by alcohol last updated on 25/Nov/21 $$\mathrm{1}+\mathrm{4}+\frac{\mathrm{16}}{\mathrm{2}}+\frac{\mathrm{64}}{\mathrm{6}}+…+\frac{\mathrm{4}^{{n}} }{{n}!}=? \\ $$ Answered by puissant last updated on 25/Nov/21 $${S}_{{n}} =\mathrm{1}+\mathrm{4}+\frac{\mathrm{16}}{\mathrm{2}}+\frac{\mathrm{64}}{\mathrm{6}}+….+\frac{\mathrm{4}^{{n}} }{{n}!}\:=\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\mathrm{4}^{{k}}…

let-give-u-n-k-1-n-a-k-2-with-a-k-sequence-of-reals-a-k-gt-0-and-v-n-k-1-n-a-k-k-prove-that-u-n-converges-v-n-converges-

Question Number 28977 by abdo imad last updated on 02/Feb/18 $${let}\:{give}\:{u}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:{a}_{{k}} ^{\mathrm{2}} \:\:\:\:\:{with}\:\left({a}_{{k}} \right)\:{sequence}\:{of}\:{reals}/{a}_{{k}>\mathrm{0}} \\ $$$${and}\:{v}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{a}_{{k}} }{{k}}\:.\:{prove}\:{that}\:{u}_{{n}} {converges}\Rightarrow\left({v}_{{n}} \right){converges}…