All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 194241 by tri26112004 last updated on 01/Jul/23
Answered by mr W last updated on 02/Jul/23
assumeda,b∈N1n(n+a)(n+b)=An+Bn+a+Cn+b(A+B+C)n2+[(2a+b)A+bB]n+abA=1A+B+C=0(2a+b)A+bB=0abA=1⇒A=1ab⇒B=−2a+bab2⇒C=2b2∑∞n=11n(n+a)(n+b)=∑∞n=1(An+Bn+a+Cn+b)=∑∞n=1(An)+∑∞n=1(Bn+a)+∑∞n=1(Cn+b)=∑bn=1(An)+∑b−an=1(Bn+a)+∑∞n=b+1(An)+∑∞n=b−a+1(Bn+a)+∑∞n=1(Cn+b)=∑bn=1(An)+∑b−an=1(Bn+a)+∑∞n=b+1(An)+∑∞n=b+1(Bn)+∑∞n=b+1(Cn)=∑bn=1(An)+∑b−an=1(Bn+a)+∑∞n=b+1(A+B+Cn)=∑bn=1(An)+∑b−an=1(Bn+a)=∑bn=1(An)+∑bn=a+1(Bn)=1ab[∑bn=11n−(2ab+1)∑bn=a+11n]=1ab(∑an=11n−2ab∑bn=a+11n)=1ab∑an=11n−2b2∑bn=a+11n✓
Commented by tri26112004 last updated on 03/Jul/23
Thankyousomuch
Terms of Service
Privacy Policy
Contact: info@tinkutara.com