Question 1059746
I'll find the first term.
You do the rest.
{{{a[1]=2}}}
{{{a[n+1]=a[n]-1/n}}}
So,
{{{a[1+1]=a[1]-1/1}}}
{{{a[2]=a[1]-1}}}
{{{a[2]=2-1}}}
{{{a[2]=1}}}
So then,
{{{a[2+1]=a[2]-1/2}}}
{{{a[3]=1-1/2}}}
Finish that and then find the next two similarly.