Q-158528-P-n-1-n-1-3-1-n-1-3-1-P-n-1-n-1-3-1-3-n-1-3-1-3-P-n-1-n-1-1-n-2-2n-1-n-1-1-n-1-1-n-2-2n-1-n-1-1-P-n-1- Tinku Tara June 4, 2023 Mensuration 0 Comments FacebookTweetPin Question Number 158751 by puissant last updated on 08/Nov/21 Q158528P=∏∞n=1((n+1)3−1(n+1)3+1)⇒P=∏∞n=1((n+1)3−13(n+1)3+13)⇒P=∏∞n=1{(n+1−1)(n2+2n+1+n+1+1)(n+1+1)(n2+2n+1−n−1+1)}⇒P=∏∞n=1{nn+2}∙∏∞n=1{n2+3n+3n2+n+1}=limn→∞∏nk=1{kk+2}∙limn→∞∏nk=1{k2+3k+3k2+k+1}=limn→∞{13∙24∙35∙…∙nn+2}×limn→∞{73∙137∙…∙n2+3n+3n2+n+1}=2limn→∞{1(n+1)(n+2)}×13limn→∞{n2+3n+3}=23limn→∞{n2+3n+3n2+3n+2}=23limn→∞{1+3n+3n21+3n+2n2}=23.P=∏∞n=1((n+1)3−1(n+1)3+1)=23..……………Lepuissant…………… Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-93208Next Next post: Find-square-root-of-7-30-2-i- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.