Can someone help with these questions.

Question 1)

The method from the previous question can also be used for finite telescoping sums. For example: 1 1 H + + +...+ Number 1 x2 2 x 3 3 x4 2014 x 2015 and 1 1 1 Number 1 x 3 + 3 x ! + + ...+ 5 x 7 99 x 101By the difference of two squares formula (n + 1)2 - n2 = 2*n+1 It follows that 1+3+5+ . . . + 1001 = (12 -02) +(22 -12) + ...+ (5012 -5002) Number Say instead that we wish to calculate the following sum: 2+ 4+...+ 558, we can't use the method we used before, as that relied on the numbers being odd. However, since 2 = 1 + 1, and 4 =3 + 1 and so on, we can write: 2+4+ . ..+ 558 - (1+1) + (3 + 1) + (5 + 1) + ... + (557 + 1) So we can evaluate this sum: 2+4+...+ 558 NumberLet's use partial fractions and telescoping series to evaluate: k=1 12+ 3 k+ 2 By using partial fractions and a telescoping sum, we find that n 1 Sn = k=1 k2 + 3k+2 Thus the original infinite sum is lim sn = Number 12+ 3k+ 2 n->OO k=1Consider the sequence \"k = (1)k. The sum of the rst in; terms of this sequence 3-11 = (1)1 + (1}2 + . . . +(1}n' - equals' 0 l when n is even, and - equals ' 1 I when n is odd. SO 3\" is' convergent v ' . But if you still feel this sequence should have a limiting value, we can consider the average of the partial sums k=1 Now, A\" - equals ' In . when n is even, and - equals ' In - when n is odd. Then Hm An 2' Number I .This is called the Cesare sum of the original (1k sequence. ni-OO Note that there is a song called Cesaro Summability, but we don't know why. if 11in 3?: = c then the Cesaro sum will also converge to c. n no ,, Cesaro summation allows us to assign a value to some non-convergent series (but not always)