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Unformatted text preview: k = 2 ` + 1 is odd, we have ( k !) 1 /k = (1 Â· 2 Â· 3 Â·Â·Â· ` Â· ( ` + 1) Â·Â·Â· k )  {z } k = 2 ` + 1 factors 1 /k â‰¥ (( ` + 1) Â·Â·Â· k )  {z } ` + 1 factors 1 /k â‰¥ ( ` + 1) ` +1 k â‰¥ ( 1 2 k ) 1 2 (since ` + 1 = k +1 2 > 1 2 k â‡’ ` + 1 k > 1 2 ) In either case, ( k !) 1 /k â‰¥ ( 1 2 k ) 1 2 â†’ âˆž as k â†’ âˆž . Thus r = 0 and we have convergence. example 1.3. See whether or not the the root test applies to the series âˆž X k =1 1 2 k +(1) k . solution. We have  a k  1 /k = Â± Â± Â± Â± 1 2 k +(1) k Â± Â± Â± Â± 1 /k = 1 2 ( k +(1) k ) /k = 1 2 1+ 1 k (1) k â†’ 1 2 as k â†’ âˆž Since lim k â†’ âˆž  a k  1 /k = 1 2 < 1 the series converges. 2...
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 Fall '08
 HIMONAS
 Calculus, Mathematical Series, 1 Â K, 1 2K, 1 K

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