Sum of the series

**thereddevils** · February 8th 2009, 12:55 AM

Write down in full the sum of each of the following series , and hence find its value .

(1) $\sum^{5}_{r=1}\frac{90}{r}$

(2) $\sum^{8}_{r=1}\sin\frac{r\pi}{3}$

(3) $\sum^{5}_{r=0}(-1)^r(1+2^{r+1})$

My thought process :

(1) $90\sum^{5}_{r=1}\frac{1}{r}$
$=90\frac{2}{n(n+1)}$
$=90\frac{2}{5(6)}$
=6 ---- my ans is wrong .

I have no idea how to do the rest .

**elliotyang** · February 8th 2009, 01:15 AM

well, if you cannot think of the general of the sum, then use common sense. since the nth number is quite less, we can substitute the value one by one.
for example, for first question, it will be
90(1/1 + 1/2+ 1/3+ 1/4+ 1/5)

**thereddevils** · February 8th 2009, 02:05 AM

I am sorry for not including this in my previous post . How to apply the formulas to find the sum of these series instead of listing down all the terms and adding them up ?

**mr fantastic** · February 8th 2009, 02:44 AM

Originally Posted by thereddevils

Write down in full the sum of each of the following series , and hence find its value .

(1) $\sum^{5}_{r=1}\frac{90}{r}$

(2) $\sum^{8}_{r=1}\sin\frac{r\pi}{3}$

(3) $\sum^{5}_{r=0}(-1)^r(1+2^{r+1})$

My thought process :

(1) $90\sum^{5}_{r=1}\frac{1}{r}$
$=90\frac{2}{n(n+1)}$
$=90\frac{2}{5(6)}$
=6 ---- my ans is wrong .

I have no idea how to do the rest .

Of course your answer is wrong. $\sum^{n}_{r=1}\frac{1}{r} \neq \frac{1}{\sum^{n}_{r=1} r}$ . eg. $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \neq \frac{1}{1 + 2 + 3 + 4 + 5}$ should be obvious.

Just write each sum out in full and then add up using basic arithmetic.

Originally Posted by thereddevils

I am sorry for not including this in my previous post . How to apply the formulas to find the sum of these series instead of listing down all the terms and adding them up ?

There is certainly no formula for the sum of reciprocals. The fact that there are only a few terms in each of your sums should be the clue that you write the sum out in full and then calculate.

**thereddevils** · February 8th 2009, 05:47 AM

Thanks . Is it the same for (2) and (3) ?

**mr fantastic** · February 8th 2009, 05:36 PM

Originally Posted by thereddevils

Thanks . Is it the same for (2) and (3) ?

Yes.

**elliotyang** · February 8th 2009, 05:40 PM

yea, the method is the same. for example,
(2) would be:
sin(pi/3)+sin(2pi+3)+...+sin(8pi/3)

**oswaldo** · February 8th 2009, 06:31 PM

$<br /> \sum^{8}_{r=1}\sin\frac{r\pi}{3}=\sin\frac{\pi}{3} +\sin\frac{2\pi}{3}+\ldots+\sin\frac{8\pi}{3}=\sin \frac{7\pi}{3}+\sin\frac{8\pi}{3}=2 \sin\frac{\pi}{3}=\sqrt{3}$

Because:
$<br /> \sin\frac{r\pi}{3}= - \sin\frac{(r+3)\pi}{3}$

-O

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