Find $lim_{nto infty}sum_{k=1}^{n}(sinfrac{pi}{2k}-cosfrac{pi}{2k}-sinfrac{pi}{2(k+2)}+cosfrac{pi}{2(k+2)})$
$begingroup$
Find $lim_{nto infty}sum_{k=1}^{n}left(sinfrac{pi}{2k}-cosfrac{pi}{2k}-sinfrac{pi}{2(k+2)}+cosfrac{pi}{2(k+2)}right)$
$$lim_{nto infty}sum_{k=1}^{n}left(sinfrac{pi}{2k}-cosfrac{pi}{2k}-sinfrac{pi}{2(k+2)}+cosfrac{pi}{2(k+2)}right)$$
$$lim_{nto infty}sum_{k=1}^{n}left(sinfrac{pi}{2k}-sinfrac{pi}{2(k+2)}-cosfrac{pi}{2k}+cosfrac{pi}{2(k+2)}right)$$
$$lim_{nto infty}sum_{k=1}^{n}left(2cosfrac{(frac{pi}{2k}+frac{pi}{2(k+2)})}{2}sinfrac{(frac{pi}{2k}-frac{pi}{2(k+2)})}{2}+2sinfrac{(frac{pi}{2k}+frac{pi}{2(k+2)})}{2}sinfrac{(frac{pi}{2k}-frac{pi}{2(k+2)})}{2}right)$$
I am stuck here.
limits trigonometry definite-integrals
edited Dec 27 '18 at 16:19
Namaste
1
asked Dec 27 '18 at 15:34
user984325user984325
246112
$endgroup$
add a comment |
$begingroup$
Find $lim_{nto infty}sum_{k=1}^{n}left(sinfrac{pi}{2k}-cosfrac{pi}{2k}-sinfrac{pi}{2(k+2)}+cosfrac{pi}{2(k+2)}right)$
$$lim_{nto infty}sum_{k=1}^{n}left(sinfrac{pi}{2k}-cosfrac{pi}{2k}-sinfrac{pi}{2(k+2)}+cosfrac{pi}{2(k+2)}right)$$
$$lim_{nto infty}sum_{k=1}^{n}left(sinfrac{pi}{2k}-sinfrac{pi}{2(k+2)}-cosfrac{pi}{2k}+cosfrac{pi}{2(k+2)}right)$$
$$lim_{nto infty}sum_{k=1}^{n}left(2cosfrac{(frac{pi}{2k}+frac{pi}{2(k+2)})}{2}sinfrac{(frac{pi}{2k}-frac{pi}{2(k+2)})}{2}+2sinfrac{(frac{pi}{2k}+frac{pi}{2(k+2)})}{2}sinfrac{(frac{pi}{2k}-frac{pi}{2(k+2)})}{2}right)$$
I am stuck here.
limits trigonometry definite-integrals
edited Dec 27 '18 at 16:19
Namaste
1
asked Dec 27 '18 at 15:34
user984325user984325
246112
$endgroup$
add a comment |
$begingroup$
Find $lim_{nto infty}sum_{k=1}^{n}left(sinfrac{pi}{2k}-cosfrac{pi}{2k}-sinfrac{pi}{2(k+2)}+cosfrac{pi}{2(k+2)}right)$
$$lim_{nto infty}sum_{k=1}^{n}left(sinfrac{pi}{2k}-cosfrac{pi}{2k}-sinfrac{pi}{2(k+2)}+cosfrac{pi}{2(k+2)}right)$$
$$lim_{nto infty}sum_{k=1}^{n}left(sinfrac{pi}{2k}-sinfrac{pi}{2(k+2)}-cosfrac{pi}{2k}+cosfrac{pi}{2(k+2)}right)$$
$$lim_{nto infty}sum_{k=1}^{n}left(2cosfrac{(frac{pi}{2k}+frac{pi}{2(k+2)})}{2}sinfrac{(frac{pi}{2k}-frac{pi}{2(k+2)})}{2}+2sinfrac{(frac{pi}{2k}+frac{pi}{2(k+2)})}{2}sinfrac{(frac{pi}{2k}-frac{pi}{2(k+2)})}{2}right)$$
I am stuck here.
limits trigonometry definite-integrals
edited Dec 27 '18 at 16:19
Namaste
1
asked Dec 27 '18 at 15:34
user984325user984325
246112
$endgroup$
Find $lim_{nto infty}sum_{k=1}^{n}left(sinfrac{pi}{2k}-cosfrac{pi}{2k}-sinfrac{pi}{2(k+2)}+cosfrac{pi}{2(k+2)}right)$
$$lim_{nto infty}sum_{k=1}^{n}left(sinfrac{pi}{2k}-cosfrac{pi}{2k}-sinfrac{pi}{2(k+2)}+cosfrac{pi}{2(k+2)}right)$$
$$lim_{nto infty}sum_{k=1}^{n}left(sinfrac{pi}{2k}-sinfrac{pi}{2(k+2)}-cosfrac{pi}{2k}+cosfrac{pi}{2(k+2)}right)$$
$$lim_{nto infty}sum_{k=1}^{n}left(2cosfrac{(frac{pi}{2k}+frac{pi}{2(k+2)})}{2}sinfrac{(frac{pi}{2k}-frac{pi}{2(k+2)})}{2}+2sinfrac{(frac{pi}{2k}+frac{pi}{2(k+2)})}{2}sinfrac{(frac{pi}{2k}-frac{pi}{2(k+2)})}{2}right)$$
I am stuck here.
limits trigonometry definite-integrals
limits trigonometry definite-integrals
edited Dec 27 '18 at 16:19
Namaste
1
asked Dec 27 '18 at 15:34
user984325user984325
246112
edited Dec 27 '18 at 16:19
Namaste
1
asked Dec 27 '18 at 15:34
user984325user984325
246112
edited Dec 27 '18 at 16:19
Namaste
1
edited Dec 27 '18 at 16:19
Namaste
1
edited Dec 27 '18 at 16:19
Namaste
1
1
asked Dec 27 '18 at 15:34
user984325user984325
246112
asked Dec 27 '18 at 15:34
user984325user984325
246112
asked Dec 27 '18 at 15:34
user984325user984325
246112
246112
add a comment |
add a comment |
1 Answer
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$begingroup$
Hint. One may recall teslescoping sums, by writing
$$
u_{k+2}-u_k=left(u_{k+2}-u_{k+1} right)+left(u_{k+1}-u_{k} right)
$$
giving
$$
sum_{k=1}^nleft(u_{k+2}-u_{k} right)=left(u_{n+2}-u_2right)+left(u_{n+1}-u_1right).
$$
answered Dec 27 '18 at 15:43
Olivier OloaOlivier Oloa
109k17178294
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
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active
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active
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$begingroup$
Hint. One may recall teslescoping sums, by writing
$$
u_{k+2}-u_k=left(u_{k+2}-u_{k+1} right)+left(u_{k+1}-u_{k} right)
$$
giving
$$
sum_{k=1}^nleft(u_{k+2}-u_{k} right)=left(u_{n+2}-u_2right)+left(u_{n+1}-u_1right).
$$
answered Dec 27 '18 at 15:43
Olivier OloaOlivier Oloa
109k17178294
$endgroup$
add a comment |
$begingroup$
Hint. One may recall teslescoping sums, by writing
$$
u_{k+2}-u_k=left(u_{k+2}-u_{k+1} right)+left(u_{k+1}-u_{k} right)
$$
giving
$$
sum_{k=1}^nleft(u_{k+2}-u_{k} right)=left(u_{n+2}-u_2right)+left(u_{n+1}-u_1right).
$$
answered Dec 27 '18 at 15:43
Olivier OloaOlivier Oloa
109k17178294
$endgroup$
add a comment |
$begingroup$
Hint. One may recall teslescoping sums, by writing
$$
u_{k+2}-u_k=left(u_{k+2}-u_{k+1} right)+left(u_{k+1}-u_{k} right)
$$
giving
$$
sum_{k=1}^nleft(u_{k+2}-u_{k} right)=left(u_{n+2}-u_2right)+left(u_{n+1}-u_1right).
$$
answered Dec 27 '18 at 15:43
Olivier OloaOlivier Oloa
109k17178294
$endgroup$
Hint. One may recall teslescoping sums, by writing
$$
u_{k+2}-u_k=left(u_{k+2}-u_{k+1} right)+left(u_{k+1}-u_{k} right)
$$
giving
$$
sum_{k=1}^nleft(u_{k+2}-u_{k} right)=left(u_{n+2}-u_2right)+left(u_{n+1}-u_1right).
$$
answered Dec 27 '18 at 15:43
Olivier OloaOlivier Oloa
109k17178294
edited Dec 31 '18 at 7:52
answered Dec 27 '18 at 15:43
Olivier OloaOlivier Oloa
109k17178294
answered Dec 27 '18 at 15:43
Olivier OloaOlivier Oloa
109k17178294
answered Dec 27 '18 at 15:43
Olivier OloaOlivier Oloa
109k17178294
109k17178294
add a comment |
add a comment |
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