For every k ∈ N , let x k = ∑ n = 1 ∞ 1 n 2 ( 1 − 1 2 n + 1 4 n 2 ) 2 k .

Question

For every   k  ∈      N  , let      x    k    =      ∑          n      =      1              ∞            1          n      2                  (      1      −              1                  2          n                    +              1                  4                      n            2                              )              2      k        .

escampetaq5 · Accepted Answer

Solution 1: By the dominated convergence theorem, we may switch the order of the limit and the sum, so we see that      lim          k      →      ∞            x    k    =  0.Solution 2: Notice the terms are always bounded above by       1          n      2      . Let   ϵ  &amp;gt;  0, and choose N such that      ∑          n      =      N        ∞        1          n      2        &amp;lt;  ϵ  .Then choose k so large that            (      1      −              1                  2          N                    )              2      k        ≤      ϵ    N    .It then follows that       |        x    k        |    ≤  2  ϵ  , and the same inequality holds for all   j  ≥  k. Since   ϵ was arbitrary the proof is finished.

For every k ∈ <mrow class="MJX-TeXAtom-ORD"> <mi mathvariant="double-str

Answered question

Answer & Explanation

New Questions in Calculus 2