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→∞.


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Question Number 408 by 112358 last updated on 30/Dec/14
$Is the sequence {u_n } defined by the formula u_n =((n+(−1.5)^n )/(lgn)) , where n∈Z^+ ∣n≥2 , convergent? If so find its sum as n→∞.$
$I s t h e s e q u e n c e {u_{n}} d e f i n e d b y t h e f o r m u l a u_{n} = \frac{n + {(- 1.5)}^{n}}{l g n}, w h e r e n \in Z^{+} ∣ n ⩾ 2, c o n v e r g e n t ? I f s o f i n d i t s s u m a s n \to \infty .$
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

$u_{n} = \frac{n + {(- 1.5)}^{n}}{\lg n}$ $u_{2 k} = \frac{2 k + {(1.5)}^{2 k}}{\lg 2 k}$ $u_{2 k + 1} = \frac{(2 k + 1) - {(1.5)}^{2 k + 1}}{\lg (2 k + 1)}$ $We can define a sequence with i^{t h} element v_{i}$ $given by i ⩾ 1$ $v_{i} = u_{2 i} + u_{2 i + 1}$ $v_{i} = \frac{2 i}{\lg 2 i} + \frac{2 i + 1}{\lg (2 i + 1)} + \frac{{(1.5)}^{2 i}}{\lg 2 i} - \frac{{(1.5)}^{2 i + 1}}{\lg (2 i + 1)}$
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