1. ## series and sequence

for question number 10, Given Ur =2/(r+1)(r+3),and I have found that sum of first n term is ∑Ur=(5/6)-(1/n+2)-(1/n+3). Then, i found the Sum of infinty of Ur=5/6.
But i cant find the last question for ∑Ur+1 + 1/2 Anybody know this?

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2. ## Re: series and sequence

Originally Posted by goldsomecriss
for question number 10, Given Ur =2/(r+1)(r+3),and I have found that sum of first n term is ∑Ur=(5/6)-(1/n+2)-(1/n+3). Then, i found the Sum of infinty of Ur=5/6.
But i cant find the last question for ∑Ur+1 + 1/2 Anybody know this?
Is it the case that $\Large{\sum\limits_{r = 1}^\infty {\left( {{u_r} + \frac{1}{{{2^r}}}} \right)} = \sum\limits_{r = 1}^\infty {\left( {{u_r}} \right)} + \sum\limits_{r = 1}^\infty {\left( {\frac{1}{{{2^r}}}} \right)}~?}$

Is the last sum geometric?

3. ## Re: series and sequence

Originally Posted by Plato
Is it the case that $\Large{\sum\limits_{r = 1}^\infty {\left( {{u_r} + \frac{1}{{{2^r}}}} \right)} = \sum\limits_{r = 1}^\infty {\left( {{u_r}} \right)} + \sum\limits_{r = 1}^\infty {\left( {\frac{1}{{{2^r}}}} \right)}~?}$

Is the last sum geometric?
erm..i think geometric method is not involved as this is method of differences.....also, we need to find Ur+1 not Ur

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4. ## Re: series and sequence

Originally Posted by goldsomecriss
erm..i think geometric method is not involved as this is method of differences.....also, we need to find Ur+1 not Ur
You don't understand series and I do. See here.

$\large{\sum\limits_{r = 1}^\infty {\left( {\dfrac{1}{{{2^r}}}} \right)} = 1~\&~ \dfrac{5}{6} + 1 = \dfrac{{11}}{6}}$

5. ## Re: series and sequence

Originally Posted by Plato
You don't understand series and I do. See here.

$\large{\sum\limits_{r = 1}^\infty {\left( {\dfrac{1}{{{2^r}}}} \right)} = 1~\&~ \dfrac{5}{6} + 1 = \dfrac{{11}}{6}}$
thanks for giving some ideas, but the answer given is (19/12)....also, what i saw here is (1/2∧r) =1 ,how could it be?...

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6. ## Re: series and sequence

Originally Posted by goldsomecriss
thanks for giving some ideas, but the answer given is (19/12)....also, what i saw here is (1/2∧r) =1 ,how could it be?...
A geometric series $\sum\limits_{k = J}^\infty {\left( {{r^k}} \right)}=\dfrac{r^J}{1-r}$ provided $|r|<1$

Therefore if $r=\dfrac{1}{2}$ then $\sum\limits_{k = 1}^\infty {\left( {{r^k}} \right)}=\dfrac{2^{-1}}{1-2^{-1}}=1$

You rightly observed that for $u_r=\dfrac{2}{(r+1)(x+3)}$ then $\sum\limits_{k = 1}^\infty {\left( {{u_k}} \right) = } \frac{{11}}{6}$.

Now we are asked to find $\sum\limits_{k = 1}^\infty {\left( {{u_{k + 1}} + \frac{1}{{{2^r}}}} \right)}$
Note that $\sum\limits_{k = 1}^\infty {\left( {{u_{k + 1}}} \right)}=\sum\limits_{k = 1}^\infty {\left( {{u_{k }}} \right)}-u_1$ AND $u_1=\frac{1}{4}$.

Does $\dfrac{{11}}{6}-\dfrac{{1}}{4}=\dfrac{{19}}{12}~?$ That is no accident.

7. ## Re: series and sequence

Originally Posted by goldsomecriss
thanks for giving some ideas, but the answer given is (19/12)....also, what i saw here is (1/2∧r) =1 ,how could it be?...

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No, $\frac{1}{2^r}$ is not equal to 1! $\sum_{r= 1}^\infty \frac{1}{2^r}= \frac{1}{2}+ \frac{1}{4}+ \frac{1}{8}+ \cdot\cdot\cdot= 1$

Notice that 1/2+ 1/4= 3/4= 1- 1/4, 1/2+ 1/4+ 1/8= 3/4+ 1/8= 6/8+ 1/8= 7/8= 1- 1/8, 1/2+ 1/4+ 1/8+ 1/16= 7/8+ 1/16= 14/16= 1/16= 15/16= 1- 1/16, etc. getting closer and closer to 1. In fact, each new term is exactly half the distance from the previous sum to 1/2.

8. ## Re: series and sequence

oh...thank you so much!

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9. ## Re: series and sequence

Originally Posted by Plato
A geometric series $\sum\limits_{k = J}^\infty {\left( {{r^k}} \right)}=\dfrac{r^J}{1-r}$ provided $|r|<1$

Therefore if $r=\dfrac{1}{2}$ then $\sum\limits_{k = 1}^\infty {\left( {{r^k}} \right)}=\dfrac{2^{-1}}{1-2^{-1}}=1$

You rightly observed that for $u_r=\dfrac{2}{(r+1)(x+3)}$ then $\sum\limits_{k = 1}^\infty {\left( {{u_k}} \right) = } \frac{{11}}{6}$.

Now we are asked to find $\sum\limits_{k = 1}^\infty {\left( {{u_{k + 1}} + \frac{1}{{{2^r}}}} \right)}$
Note that $\sum\limits_{k = 1}^\infty {\left( {{u_{k + 1}}} \right)}=\sum\limits_{k = 1}^\infty {\left( {{u_{k }}} \right)}-u_1$ AND $u_1=\frac{1}{4}$.

Does $\dfrac{{11}}{6}-\dfrac{{1}}{4}=\dfrac{{19}}{12}~?$ That is no accident.
oh..ok,get it, thank you so much

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