Is-the-sequence-u-n-defined-by-the-formula-u-n-n-1-5-n-lgn-where-n-Z-n-2-convergent-If-so-find-its-sum-as-n-

Question Number 408 by 112358 last updated on 30/Dec/14

I s t h e s e q u e n c e {u_{n}} d e f i n e d b y t h e f o r m u l a u_{n} = \frac{n + {(- 1.5)}^{n}}{l g n}, w h e r e n \in Z^{+} ∣ n ⩾ 2, c o n v e r g e n t ? I f s o f i n d i t s s u m a s n \to \infty .

Commented by 123456 last updated on 30/Dec/14

u_{n} = \frac{n + {(- 1.5)}^{n}}{\lg n}, n \in N, n ⩾ 2

Commented by 123456 last updated on 30/Dec/14

\frac{\infty \pm \infty}{\infty}

\frac{?}{\infty}, \frac{\infty}{\infty}

Commented by prakash jain last updated on 31/Dec/14

u_{n} = \frac{n + {(- 1.5)}^{n}}{\lg n}

u_{2 k} = \frac{2 k + {(1.5)}^{2 k}}{\lg 2 k}

u_{2 k + 1} = \frac{(2 k + 1) - {(1.5)}^{2 k + 1}}{\lg (2 k + 1)}

We can define a sequence with i^{t h} element v_{i}

given by i ⩾ 1

v_{i} = u_{2 i} + u_{2 i + 1}

v_{i} = \frac{2 i}{\lg 2 i} + \frac{2 i + 1}{\lg (2 i + 1)} + \frac{{(1.5)}^{2 i}}{\lg 2 i} - \frac{{(1.5)}^{2 i + 1}}{\lg (2 i + 1)}