Question #103518
Evaluate using root test.

Infinity

Summation sign (1- 1/k)^k

k = 1
1
Expert's answer
2020-02-23T15:11:13-0500

limk((11/(k+1))k+1/(11/k)k)lim _{k\to \infty}((1-1/(k+1))^{k+1}/(1-1/k)^k)

limk((k/(k+1))k+1/((k1)/k)k)lim _{k\to \infty}((k/(k+1))^{k+1}/((k-1)/k)^k)

limk((k2/(k21))k)lim _{k\to \infty}((k^2/(k^2-1))^{k})

limk((1/(11/k2))k)lim _{k\to \infty}((1/(1-1/k^2))^{k}) is of the form 11^{\infty}

elimk(((k2/(k21))1)k)e^{lim _{k\to \infty}(((k^2/(k^2-1))-1)k)}

elimk(k/(k21))e^{lim_{k \to \infty}(k/(k^2-1))}

e0=1e^0=1

Root test fails as limit value is equal to 1.


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