Question 549757
{{{a[n]=n/(2n+1)}}} <--> {{{a[n]=1/2-1/2(2n+1)}}} with {{{a[1]=1/3}}} and increasing
{{{a[n]=(n+1)/(2n+1)}}} <--> {{{a[n]=1/2+1/2(2n+1)}}} with {{{a[1]=2/3}}} and decreasing
{{{a[n]=1/2+(1/2)^(n+1))}}} with {{{a[1]=1/2+1/4=3/4}}} if you want it to converge faster