If y n > 0, lim x n y n = a and ∑ y...

Reed Eaton

Reed Eaton

Answered

2022-06-27

If y n > 0, lim x n y n = a and y n = + , then lim x 1 + x 2 + . . . + x n y 1 + y 2 + . . . + y n = a

Answer & Explanation

Trey Ross

Trey Ross

Expert

2022-06-28Added 30 answers

You had a good idea. Indeed, since x n y n a, then for any ε > 0 there is some N such that for k N you get ( a ε ) y k x k ( a + ε ) y k (we've used that y k > 0 here, so that the signs of inequalities remain). Then for n N you can rewrite
x 1 + . . . + x n y 1 + . . . + y n = x 1 + . . . + x N y 1 + . . . + y n + x N + 1 + . . . x N y 1 + . . . + y n x 1 + . . . + x N y 1 + . . . y n + ( a + ε ) y N + 1 + . . . + y n y 1 + . . . + y n
Now, with N fixed, you have x 1 + . . . + x N y 1 + . . . + y n 0 as n , due to divergence of series k y k . Moreover,
y N + 1 + . . . y n y 1 + . . . + y n = 1 y 1 + . . . + y N y 1 + . . . + y n n 1 0 = 1
due to the same reason. Hence lim sup n x 1 + . . . + x n y 1 + . . . + y n a + ε. Similarly you can prove that lim inf n x 1 + . . . x n y 1 + . . . + y n a ε and since ε > 0 was arbitrary you can conclude
a lim inf n x 1 + . . . + x n y 1 + . . . + y n lim sup n x 1 + . . . + x n y 1 + . . . + y n a
and so the limit exists and is equal to a.

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