(a)_(m)+(b)_(1)+(b)_(2)+... +(b)_(n)}(m+n)=-|||-dfrac (m)(m+n)cdot dfrac ({a)_(1
+dfrac (sin n)({2)^n} ;-|||-(2) _(n)=1+dfrac (1)({2)^2}+dfrac (1)({3)^2}+... +df
(4) lim _(narrow infty )((1+dfrac {2)(n)+dfrac (2)({n)^2})}^n.
+(n)^3);-|||-(2) lim _(narrow infty )n[ dfrac (1)({(n+1))^2}+dfrac (1)({(n+2))^2
lim _(narrow infty )(dfrac (1)({n)^2+n+1}+dfrac (2)({n)^2+n+2}+... +dfrac (n)({n
+dfrac (sin npi )(n+dfrac {1)(n)}]
+dfrac (n)({n)^2+n+n})=dfrac (1)(2)证明:
+dfrac (1)({2)^n})-|||-1/2^n);(12)-|||-(13) lim _(narrow infty )dfrac ((n+1)(n+2
,求幂级数 sum _(n=1)^infty dfrac (2n-1)({2)^n}(x)^2n-2 的和函数,并求级数 sum _(n=1)^infty df
+dfrac (1)(n(n+1)) =-|||-(3) lim _(narrow infty )(dfrac (1)(2)+dfrac (3)({2)^2}+