2. Consider the alternating series 1 –+-+..., i.e. Eak where az = (-1)k-1 for all k e N. 4 Let Sn =E- ak denote the partial sums. (a) Show that Eak is not absolutely convergent. (b) Show that for all n E N, 1 San =E 1 = S2n + and S2n+1 2k(2k – 1) 2n + 1 k=1 (Hint: group pairs of odd and even terms.) (c) Prove that the sequence of even partial sums (S2n) has a limit S. (d) Prove that the whole sequence (Sm) has the same limit S, and hence , ak = S.

Q: 3y² - 2y-1 (-)-5y²- 2y +6

A:

Q: Evaluate the double integral. dA, x2 + 1 {(x, y) | 0 < x < 2, 0 < y < Vx}

A: We are clearly given the limits of the integral. Since y limits are in terms of x so dA = dy dx

Q: 2) 36³x=¹

A:

Q: expand (a-b)7

A: The solution is given in the next step.

Q: 4root2*4root2

A:

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A: The expression is given as 1+–3m+–9m+4+5+–10m. Simplifying the given expression. 1+–3m+–9m+4+5+–10m…

Q: 4. Expand and simplify (2x + 3)*

A: We have to expand and simplify the following expression: . We will apply the binomial theorem to…

Q: d. 1 3a+18 = 1 - 8 a+6

A:

Q: limx→- f(x) limo+ f(x) limx→ = = =

A:

Q: 18(9+7)+3

A: 18(9+7)+3

Q: expand (x+y)6

A:

Q: limx→2 f(x) = lim, ¹x→−1+ f(x) =

A: Given graph of the function fx is   1. Find :  limx→2 fx. 2. Find :limx→-1+ fx.

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A: 3(m-5)+m=0 3m-15+m=0 add both side 15 we get 2m=15

Q: -5 4 -2 -7 -8 2.

A: