Question Number 8661 by FilupSmith last updated on 20/Oct/16
$${p}_{{n}} ={n}^{{th}} \:\mathrm{prime}\:\mathrm{number} \\ $$$${p}_{\mathrm{1}} =\mathrm{2},\:{p}_{\mathrm{2}} =\mathrm{3},\:{p}_{\mathrm{3}} =\:\mathrm{5},\:… \\ $$$$ \\ $$$$\mathrm{Does}\:\mathrm{the}\:\mathrm{following}\:\mathrm{converge}: \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{p}_{{i}} }{{p}_{{i}+\mathrm{1}} } \\ $$$$\mathrm{Prove}/\mathrm{disprove} \\ $$
Answered by prakash jain last updated on 21/Oct/16
$${a}_{{n}} =\frac{{p}_{{n}} }{{p}_{{n}+\mathrm{1}} } \\ $$$$\mathrm{Bertrand}\:\mathrm{prime}\:\mathrm{number}\:\mathrm{theorem}\:\mathrm{states} \\ $$$$\mathrm{that}\:{p}_{{n}+\mathrm{1}} <\mathrm{2}{p}_{{n}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} >\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{p}_{{n}} }{\mathrm{2}{p}_{{n}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{Since}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} \neq\mathrm{0}.\:\mathrm{The}\:\mathrm{series}\:\mathrm{diverges}. \\ $$