Sum-to-n-terms-1-1-1-1-2-1-1-2-3-

Question Number 2273 by RasheedAhmad last updated on 12/Nov/15

S u m t o n t e r m s .

\frac{1}{1} + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} \dots

Answered by Yozzi last updated on 12/Nov/15

F o r a n A . P w i t h f i r s t t e r m a a n d

c o m m o n d i f f e r e n c e d, i t s s u m t o n

t e r m s i s g i v e n b y

S (n) = \frac{n}{2} (2 a + (n - 1) d) .

F o r A . P s i n d e m o m i n a t o r, t h e

f i r s t t e r m i s a = 1 a n d c o m m o n

d i f f e r e n c e i s d = 1 . T h u s, t h e s u m t o r

t e r m s o f e a c h A . P i n t h e d e n o m i n a t o r s,

i s g i v e n b y

S (r) = \frac{r}{2} (2 \times 1 + (r - 1) \times 1)

= \frac{r}{2} (2 + r - 1)

S (r) = \frac{r}{2} (r + 1) .

T h e g e n e r a l t e r m U_{r} f o r t h e g i v e n

s e r i e s i s

U_{r} = \frac{1}{S (r)} = \frac{2}{r (r + 1)} .

O b s e r v e t h a t

\frac{1}{r} - \frac{1}{r + 1} = \frac{r + 1 - r}{r (r + 1)} = \frac{1}{r (r + 1)} .

∴ U_{r} = 2 (\frac{1}{r} - \frac{1}{r + 1})

N o w l e t f (r) = 1 / r (r \in N)

\Rightarrow U_{r} = 2 (f (r) - f (r + 1))

∴ \underset{r = 1}{\sum^{n}} U_{r} = 2 [f (1) - f (2)] + 2 [f (2) - f (3)]

+ 2 [f (3) - f (4)] + 2 [f (4) - f (5)] + 2 [f (5)

- f (6)] + \dots + 2 [f (n - 3) - f (n - 2)] +

2 [f (n - 2) - f (n - 1)] + 2 [f (n - 1) - f (n)]

+ 2 [f (n) - f (n + 1)]

\Rightarrow \underset{r = 1}{\sum^{n}} U_{r} = 2 [f (1) - f (n + 1)] = 2 (\frac{1}{1} - \frac{1}{n + 1})

\underset{r = 1}{\sum^{n}} \frac{2}{r (r + 1)} = 2 (1 - \frac{1}{n + 1})

C o r o l l a r y :

\lim_{n \to \infty} \underset{r = 1}{\sum^{n}} \frac{2}{r (r + 1)} = \lim_{n \to \infty} [2 (1 - \frac{1}{n + 1})]

= 2 - \frac{2}{\infty}

= 2

Commented by Rasheed Soomro last updated on 13/Nov/15

\overset{VERY}{NICE}

Answered by Rasheed Soomro last updated on 13/Nov/15

\underline{AN EASY APPROACH} (Technically Simpler)

\frac{1}{1} + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \dots + \frac{1}{1 + 2 + \dots + n}

∵ 1 + 2 + \dots + n = \frac{n}{2} (n + 1)

∴ \frac{1}{\frac{1}{2} (1 + 1)} + \frac{1}{\frac{2}{2} (2 + 1)} + \dots + \frac{1}{\frac{n}{2} (n + 1)}

1 + \frac{1}{3} + \frac{1}{6} + \dots + \frac{2}{n (n + 1)}

L e t

\frac{A}{n} + \frac{B}{n + 1} = \frac{2}{n (n + 1)}

A (n + 1) + B (n) = 2

F o r n = 0, A = 2

F o r n = - 1, B = - 2

\frac{2}{n (n + 1)} = \frac{2}{n} - \frac{2}{n + 1}

= 1 + \frac{1}{3} + \frac{1}{6} + \dots + (\frac{2}{n} - \frac{2}{n + 1})

= (\frac{2}{1} - \frac{2}{1 + 1}) + (\frac{2}{2} - \frac{2}{2 + 1}) + (\frac{2}{3} - \frac{2}{3 + 1}) \dots + (\frac{2}{n} - \frac{2}{n + 1})

= \frac{2}{1} - \frac{2}{2} + \frac{2}{2} - \frac{2}{3} + \frac{2}{3} - \frac{2}{4} + \frac{2}{4} - \dots . + \frac{2}{n - 1} - \frac{2}{n} + \frac{2}{n} - \frac{2}{n + 1}

= 2 - \frac{2}{n + 1} = 2 (\frac{n + 1 - 1}{n + 1}) = \frac{2 n}{n + 1}