Question #303202

Check whether the sequence {an}, where


an= 1/(n+1)+1/(n+2)+..1/2n is convergent or not

1
Expert's answer
2022-03-01T12:12:20-0500

ANSWER. The sequence {an}\left \{ a_{n} \right \} is convergent.

EXPLANATION.

Since an=k=1n1n+ka_{n}=\sum_{k=1}^{n}\frac{1}{n+k} , then an+1=k=1n+11(n+1)+ka_{n+1}=\sum_{k=1}^{n+1}\frac{1}{(n+1) +k} and an+1an=12n+1+12n+21n+1=12n+112n+2=1(2n+1)(2n+2)>0a_{n+1} -a_{n}=\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}=\frac{1}{2n+1}-\frac{1}{2n+2} =\frac{1}{(2n+1)(2n+2)}>0 . Hence an+1>ana_{n+1}>a_{n } . So , the sequence is increasing.

Because for all kk such that 1kn1\leq k\leq n the inequality (1+n)(k+n)(1+n)\leq(k+n) is true and 1k+n11+n\frac{1}{k+n}\leq\frac{1}{1+n} . Therefore, 0<ank=1n1n+1==1n+1+...+1n+1n=nn+1<10<a_{n}\leq \sum_{k=1}^{n}\frac{1}{n+1}==\underset { n }{ \underbrace { \frac { 1 }{ n+1 } +...+\frac { 1 }{ n+1 } } }= \frac{n}{n+1} <1 .

Thus, it is proved that the increasing sequence is bounded from above, hence the sequence converges.


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