#P in NN#\{0,1};#q0#;how to solve limit?#lim_(n-oo)(qn+1)/(qn)*(qn+p+1)/(qn+p)*...*(qn+np+1)/(qn+np)#

Question

#P in NN#\{0,1};#q>0#;how to solve limit?#lim_(n->oo)(qn+1)/(qn)*(qn+p+1)/(qn+p)*...*(qn+np+1)/(qn+np)#

Paisley Johnson · Accepted Answer

See below.

#(qn+1)/(qn)*(qn+p+1)/(qn+p)cdots(qn+np+1)/(qn+np)=# #(1+1/(n q))(1+1/(nq+p))cdots(1+1/(nq+np)) =prod_(m=0)^n(1+1/(n q+p m))# Calling #L_0 = lim_(n->oo)prod_(m=0)^n(1+1/(n q+p m))# applying #log# to both sides #log(L_0)=lim_(n->oo)sum_(m=0)^n log(1+1/(n q+m p)) =# #lim_(n->oo)sum_(m=0)^nlog[(1+1/(n (q+m/n p)))^n]1/n# and this is the Riemann-Stiltjes integral #lim_(n->oo)int_(xi=0)^(xi=1) log[(1+1/(n (q+xi p)))^n]d xi# so #log(L_0)=lim_(n->oo)log((1/(1/(nq) + 1))^((nq)/p)(1 + 1/(n(p+q)))^((n(p+q))/p)((1 + n (p + q))/(1 + n q))^(1/p))# or #log(L_0)=log(e^(-1/p) e^(1/p) ((p + q)/q)^(1/p)) = log(((p + q)/q)^(1/p))# so in the end #L_0=((p + q)/q)^(1/p)#

Elijah Abram · Answer

undefined To solve the limit lim_(n->oo)(qn+1)/(qn)*(qn+p+1)/(qn+p)*...*(qn+np+1)/(qn+np), we can rewrite it as lim_(n->oo) [(qn+1)/(qn)] * [(qn+p+1)/(qn+p)] * ... * [(qn+np+1)/(qn+np)].

Using the fact that lim_(n->oo) (an+1)/(an) = lim_(n->oo) an/(an-1) = ... = lim_(n->oo) a1, where a is a constant, we can simplify the expression further.

Let's denote the given limit as L. Then, we have L = lim_(n->oo) [(qn+1)/(qn)] * [(qn+p+1)/(qn+p)] * ... * [(qn+np+1)/(qn+np)].

Now, let's focus on one term in the product, say [(qn+k+1)/(qn+k)]. We can rewrite it as [(qn+k+1)/(qn+k)] = [(qn+k+1)/(qn+k)] * [(qn+k)/(qn+k-1)] * [(qn+k-1)/(qn+k-2)] * ... * [(qn+2)/(qn+1)] * [(qn+1)/(qn)].

Notice that most of the terms in the numerator and denominator cancel out, leaving us with [(qn+k+1)/(qn)]. This cancellation happens for each term in the product.

Therefore, we can rewrite L as L = lim_(n->oo) [(qn+1)/(qn)] * [(qn+p+1)/(qn+p)] * ... * [(qn+np+1)/(qn+np)] = lim_(n->oo) [(qn+1)/(qn)] * [(qn+2)/(qn+1)] * ... * [(qn+np+1)/(qn+np)].

Now, observe that most terms in the numerator cancel out with the corresponding terms in the denominator, leaving us with [(qn+np+1)/(qn)].

Hence, the limit L simplifies to L = lim_(n->oo) [(qn+np+1)/(qn)].

To evaluate this limit, we need to consider the value of q and p. Without specific values for q and p, we cannot determine the exact value of the limit.

#P in NN#\{0,1};#q&gt;0#;how to solve limit?#lim_(n-&gt;oo)(qn+1)/(qn)*(qn+p+1)/(qn+p)*...*(qn+np+1)/(qn+np)#

#P in NN#\{0,1};#q>0#;how to solve limit?#lim_(n->oo)(qn+1)/(qn)(qn+p+1)/(qn+p)...*(qn+np+1)/(qn+np)#